Fisikastudycenter.com-Example Problem and Discussion of Heat and
Expansion, High School Grades Matter Physics 10 (X) include specific
heat, latent heat (melting heat, steam heat), heat exchange principle /
Black and expansion length principle, the volume of a material.
No Problem. 1 12 kj of heat given to the piece of metal that has a mass of 2500 grams temperature of 30 o C. If the specific heat of metal is 0.2 calories / g o C, determine the final temperature of the metal!
Discussion
Data: Q = 12 = 12000 joules kilojoules m = 2500 grams = 2.5 kg T 1 = 30 ° C c = 0.2 cal / g o C = 0.2 x 4200 joules / kg o C = 840 joules / kg o C T 2 = ...? Q = mcΔT 12000 = (2.5) (840) ΔT ΔT = 12000/2100 = 5.71 o C T 2 = T 1 + ΔT = 30 + 5.71 = 35.71 ° C No Problem. 2 500 grams of ice temperature of -12 o C is heated to a temperature of -2 ° C. If the specific heat of ice is 0.5 cal / g o C, determine much heat is needed, expressed in units of joules! Discussion Data: m = 500 grams T 1 = -12 ° C T 2 = -2 o C ΔT = T 2 - T 1 = -2 o - (-12) = 10 ° C c = 0.5 cal / g o C Q = ....? Q = mcΔT Q = (500) (0.5) (10) = 2500 calories 1 calorie = 4.2 joules Q = 2500 x 4.2 = 10500 joules No Problem. 3 500 grams of ice at 0 ° C was about to be disbursed until the entire ice into water at 0 ° C. If the specific heat of ice is 0.5 cal / g o C, and the heat of melting ice is 80 cal / g, determine much heat is needed, expressed in kilocalories! Discussion Data required: m = 500 grams L = 80 calories / gram Q = ....? Q = mL Q = (500) (80) = 40000 calories = 40 kcal No Problem. 4 500 grams of ice at 0 ° C was about to be melted into water at 5 ° C. If the specific heat of ice is 0.5 cal / g o C, heat of melting ice is 80 cal / g, and the specific heat of water 1 cal / g o C, determine much heat is needed! Discussion Data required: m = 500 grams c water = 1 cal / g o C L es = 80 calories / gram Final temperature of 5 o C → Q = .....? To make the ice 0 ° C to 5 o C to water there are two processes that must be passed: → The process of melting ice water 0 ° C to 0 ° C, the required heat called Q 1 Q 1 = mL ice = (500) (80) = 40000 calories → The process of raising the water temperature of 0 ° C to 5 ° C into water, heat is needed call the Q 2 Q 2 = mc ΔT water water = (500) (1) (5) = 2500 calories Total heat required: Q = Q 1 + Q 2 = 40000 + 2500 = 42500 calories No Problem. 5 500 grams of ice temperature was about -10 ° C until thawed into water at 5 ° C. If the specific heat of ice is 0.5 cal / g o C, heat of melting ice is 80 cal / g, and the specific heat of water 1 cal / g o C, determine much heat is needed! Discussion Data required: m = 500 grams c ice = 0.5 cal / g o C c water = 1 cal / g o C L ice = 80 cal / g Final temperature of 5 o C → Q = .....? To make the ice - 10 o C to 5 o C to water there are three processes that must be passed: → The process to raise the temperature of ice from -10 o C to ice at 0 ° C, the required heat called Q 1 Q 1 = mc ΔT es = es (500) (0.5) (10) = 2500 calories → The process of melting ice water 0 ° C to 0 ° C, the required heat called Q 2 Q 2 = mL ice = (500) (80) = 40000 calories → The process of raising the water temperature of 0 ° C to 5 ° C into water, heat is needed called Q 3 3 Q = mc ΔT water water = (500) (1) (5) = 2500 calories Total heat required: Q = Q 1 + Q 2 + Q 3 = 2500 + 40000 + 2500 = 45000 calories No Problem. 6 200 grams of water temperature of 80 o C is mixed with 300 grams of water temperature of 20 o C. Determine the temperature of the mixture? Discussion Data required: m 1 = 200 grams m 2 = 300 grams ΔT 1 = 80 - t ΔT 2 = t - 20 Final temperature = t = ......? Q = Q Thank loose ΔT m 1 c 1 1 = m 2 c 2 ΔT 2 (200) (1) (80 - t) = (300) (1) (t - 20) (2) (1) (80 - t) = (3) (1) (t - 20) 160 - 2t = 3t - 60 5t = 220 t = 44 ° C Problem # 7 100 gram piece of ice mass at 0 ° C is inserted into a cup of water with mass 200 grams temperature 50 ° C. 
If the specific heat of water is 1 cal / g ° C, ice specific heat 0.5
cal / g ° C, the heat melting the ice 80 cal / g and cups are considered
not absorb heat, what temperature is the end of a mix between ice and
water?
Discussion Question above about heat exchange / Principle Black.
Heat given off by the water used to turn himself into ice water and the
rest is used to raise the temperature of the ice that has been melting
earlier. 
with Q 1 is the heat given off water, Q 2 is the heat used to melt ice / melt and Q 3 is the heat that is used to raise the temperature of the ice has melted.
Next is an example of the problem of the mixing of hot and cold water
to take into account the heat absorbed by a vessel or container: No Problem. 8
100 g mass water temperature was 20 ° C in containers made of a
material that has a specific heat 0.20 cal / g ° C and a mass of 200 g. Then poured into a container of hot water temperature of 90 ° C as much as 800 g. If the specific heat of water is 1 cal / g ° C, determine the final temperature of a mixture of water! Discussion
Heat coming from the hot water of 90 ° C while mixing, some is absorbed
by the water temperature 20 ° and partly absorbed by the container.
There is no information related to the initial temperature of the
container, so let's just say the same temperature as the water in the
container, which is 20 ° C. Data: -Hot water m 1 = 800 g c 1 = 1 cal / g ° C Cold-water m 2 = 100 g c 2 = 1 cal / g ° C -Containers m 3 = 200 g c 3 = 0.20 cal / g ° C 
Q = Q Thank loose ΔT m 1 c 1 1 = m 2 c 2 ΔT 2 + ΔT m 3 c 3 3 
No Problem. 9 Consider the following picture! Two metal pieces are made of the same material attached. 
If the length of the metal P is twice the length of the metal Q metal thermal conductivity and P is half of the metal Q, determine the temperature at the junction between two metals? Discussion Amount of heat per unit time is equal to P metal through the heat through the metal Q. Use the formula of heat transfer by conduction: 
No Problem. 10 A steel tank that has a coefficient of expansion length of 12 x 10 -6 / ° C, and a volume of 0.05 m 3 stuffed full of gasoline that has a coefficient of expansion 950 x 10 -6 / ° C at 20 ° C. If the tank is then heated to 50 ° C, determine the volume of the spilled gasoline! (Source: Problem SNCA) Discussion 
No Problem. 11 Steel plate is heated until the temperature reaches 227 ° C to heat radiation emitted by EJ / s. If the plate continued to be heated until the temperature reaches 727 ° determine the heat radiation emitted! Discussion Data: T 1 = 227 ° C = 227 + 273 = 500 K T 2 = 727 ° C = 727 + 273 = 1000 K Heat is radiated by an object is directly proportional to the surface of the fourth power of its absolute temperature, so that: 
No Problem. 12 Stem length rail each 10 meters, installed at a temperature of 20 ° C. Expected at a temperature of 30 ° C are touching the rail. Expansion coefficient trunk railroad rails 12 × 10 -6 / ° C. The distance between the two rods are required at a temperature of 20 ° C is ... A. 3.6 mm B. 2.4 mm C. 1.2 mm D. 0.8 mm E. 0.6 mm (Problem Ebtanas 1988) Discussion
Assuming the left rail extends to the right by Δl and rail right
extends to the left of Δl, then the required gap width d is equal to
twice the Δl 
so that 
Looking dimensions of a quantity? Dimension of style? Dimension of power? Dimensions of energy? Dimension of impulse? Dimensions of momentum?
Assuming that some unknown formula X High School student, unless length
(m), mass (kg), time (s), velocity (m / s), acceleration (m / s 2), wide (2 m) and volume ( m 3) are assumed to be known, following a decrease in the amount of Physics modest dimensions.
Here is the format:
Formula ---> Unit -> Dimension
and remember:
mass -> kg -> M (from M ass time ..!?! & do kliru M ether ..!)
length -> m -> L (L ength of time ...!?!)
time -> s -> T (from T ime time ...!)
Style
mass x acceleration -> (kg) (m / s 2) -> MLT - 2
Massa Type
mass / volume -> (kg) / (m 3) -> ML -3
Energy
mass x acceleration of gravity x height -> (kg) (m / s 2) (m) -> ML 2 T - 2
Pressure
force / area ---> (kg) (m / s 2) / m 2 -> ML -1 T -2
Business
force x displacement -> (kg) (m / s 2) (m) -> ML 2 T -2
Momentum
mass x velocity -> (kg) (m / s) -> MLT -1
Impulse
force x time interval -> (kg) (m / s 2) (s) -> MLT -1
Power
Effort / time -> (kg) (m / s 2) (m) / (s) -> ML 2 T -3
Weight
mass x acceleration due to gravity -> (kg) (m / s 2) -> MLT -2
Specific gravity
weight / volume -> (kg) (m / s 2) / (m 3) -> ML -2 T -2
Some have the same amount of dimensions, such as Enterprise and Energy, Style and weight, impulse and momentum.
For a slightly more complicated matter usually displayed formula, the
brain-tweaking live, move left and right, up and down, insert the new
unit is converted to a dimension.
Just a Sample:
Formula given the force of gravity between two objects as follows
by F is the force (Newton) m 1 and m 2
are the masses of the two pieces of the object (kg), r is the distance
between the two objects (m) and G is a constant to be searched
dimensions.
From the above formula after it was found that inverted
enter the units take it to kg, m and s. To see the list above unit of force, obtained
The next example:
Given spring force equation
F = k Δ X
Where F is the spring force (Newton), Δ X is the length of the spring (meters) and k is the spring constant. Dimensional spring constant? 
Further, ..
next how to check whether or not an equation that connects certain
quantities (check formula) with analysis of two-dimensional or a formula
such as the following example: 1) The following equation linking the quantities on the motion of an object.
v t = v o + at where v t is the velocity when t, v o is the initial velocity, a is acceleration and t is time. Check with dimensional analysis of whether or not the above equation! 2) The position of an object is expressed in an equation y = At 2 + Bt + C with unit y in meters and t in second. A, B and C are constants. specify the units and dimensions of A, B and C! (Problem Fisikastudycenter) Discussion 1) The dimensions on the left side: v is the velocity t → m / s → L / T → LT -1 Dimension on the right: v is the velocity o → m / s → L / T → LT -1 at is the acceleration x time → m / s 2 xs → m / s → L / T → LT -1 Seen left side dimension equal to the dimension on the right, so the above equation is right. 2) The assumption is that the quantities are added or subtracted have the same unit or dimension with the results. Of the equation y = At 2 + Bt + C Meter ... meter = ... + ... + ... meter meter A constant unit determines Results of combination units on the meter must At 2, enter other units that have been known in this case t (time) unit is s (second) so that At 2 = m As 2 = m A = m / s 2 A dimension is LT -2 Determine the constants B unit Bt also produces meter, enter the other unit which has been known to Bt = m Bs = m B = m / s Dimension of B is LT -1 Determine the constants C unit C = m Dimension C is L
Semester Question Bank Unit and Dimension 10 SMA (Indonesian)
Fisikastudycenter.com-Example Problem and Discussion on Enterprise and Energy, Materials Physics 2 class in high school.
Include business relationships, style and movement, seek business from
the difference in kinetic energy, determine the business of the
difference in potential energy, negative positive sign on the business,
total effort, and reading graphs F - S.
No Problem. 1 A retractable beam force F = 120 N that makes an angle of 37 ° to the horizontal direction.
If the beam shift as far as 10 m, determine the work done on the block! Discussion 
No Problem. 2 Beam mass 2 kg is on slippery surfaces accelerated from rest to move with an acceleration of 2 m / s 2. 
Determine the work done on the beam during the 5 second! Discussion
First look at the beam velocity 5 seconds, then look for the difference
in the kinetic energy of the conditions of the crew and finally: 
No Problem. 3 10 kg object about to be shifted through the slippery surface of the inclined plane as shown below! 
Determine the effort required to move the object! Discussion Looking for a business with potential energy difference: 
No Problem. 4 Note the chart style (F) against displacement (S) below! 
Determine the size of the business up to 12 seconds! Discussion = The area between lines of business graph with axis S FS, for the graph above a trapezoidal area W = 1/2 (12 + 9) x 6 W = 1/2 (21) (6) W = 63 joules (Thanks tuk Rora http://r-kubik-tu-rora.blogspot.com/ above correction) No Problem. 5 A car of mass 5000 kg is moving with a speed of 72 km / h approaching a red light. 
Determine the magnitude of the braking force to do so as the car
stopped at a red light when it is 100 meters away from the car! (72 km / h = 20 m / s) Discussion 
No Problem. 6 A rod of length 40 cm and above-ground erect hammer sentenced to 10 kg from a height of 50 cm above the tip. When the average resistance force land 10 3 N, then the number of collisions hammer needs to be done in order to stick to the surface of the ground is flat .... A. 4 times B. 5 times C. 6 times D. 8 times E. 10 times (Problem UMPTN 1998) Discussion Two formulas effort involved here are:
In the hammer: W = mg Δ h
On the ground by the force of friction: W = FS Find the depth of the entry of the stick (S) by one hammer blow: FS = mgΔh (10 3) S = 10 (10) (0.5) S = 50/1000 = 5/100 m = 5 cm So once the fall of the hammer, stick into the soil as deep as 5 cm. To 40 cm long stick, then the amount of the fall of the hammer: n = 40: 5 = 8 times No Problem. 7 A block is on an incline with a friction coefficient of 0.1 as shown in the following figure. 
Beam drops down to 5 meters review. Define: a) the forces acting on the block b) the business of each force on the beam c) the total effort
Use g = 10 m / s 2, sin 53 o = 0.8, cos 53 o = 0.6, W (uppercase) for the symbol, and w (small) for the symbol gravity.
Discussion a) the forces acting on the block 
normal force (N), gravity (w) with its components, namely w sin w cos 53 ° and 53 °, the friction force F ges b) the business of each force on the beam With the incline as the trajectory (reference) displacement: Normal-Businesses by force and gravity component w cos 53 ° The second effort is zero force (force perpendicular to the track)
-Business by gravity component w sin 53 ° W = w sin 53 °. S W = mg sin 53 °. S W = (6) (10) (0.8) (5) = + 240 joules (Given a positive sign, mg sin 53 ° towards the direction of the displacement of the beam.) -Effort by friction Find great friction in advance ges f = μ N ges f = μ mg cos 53 ° ges f = (0.1) (6) (10) (0.6) = 0.36 N 3.6 N W = - fges S = - 3.6 (5) = - 18 joules (Given the negative sign, the direction opposite to the direction of the frictional force displacement of the beam) c) the total effort W total = +240 Joules - 18 joules = + 222 joules
Thanks for your submission to gita, .. the placement of a comma has been changed.
No Problem. 8 A block of mass 2 kg is on a rough inclined plane as shown in the following figure. 
Beam pushed upward by a force F = 25 N to shift up to review the extent of 5 meters. Friction that occurs between the beams with the incline of 3 N. Slope of 53 ° to the horizontal plane. Decide with positive or negative signs: a) attempt by force F b) attempt by friction c) attempt to gravity d) total effort Discussion a) attempt by force F W = F. S = + 25 (5) = + 125 joules b) attempt by friction W = - f. S = - 3 (5) = - 15 joules c) attempt to gravity W = - mg sin 53 °. S = - (2) (10) (0.8) (5) = - 80 joules d) total effort W total = + 125 - 15-80 = 30 joules
Problem # 9 Object weighing 10 N is on a slippery incline with a slope angle of 30 °. When objects slide as far as 1 m, then the work done is the gravity .... 
A. 10 sin 30 ° joule B. 10 cos 30 ° joule C. 10 sin 60 ° joule D. 10 tan 30 ° joule E. 10 tan 60 ° joule (From about Ebtanas 1990) Discussion Effort by gravity W = mg sin θ From the problem has been known that (mg) = 10 Newton θ = 30 ° and, thus W = 10 sin 30 ° joule Problem # 10 A mass of 2 kg object falls freely from the top of a high rise building 100 m. If the friction with the air and the overlooked g = 10 ms -2 then the business that is done by gravity to a height of 20 m from the ground is ..... A. 200 joules B. 400 joules C. 600 joules D. 1,600 joules E. 2,400 joules (From about Ebtanas 1992) Discussion Operations, changes in gravitational potential energy: W = mgΔ h W = 2 x 10 x (100-20) W = 1600 joules Problem # 11 A car with a mass of 1 ton moving from rest. A moment later the speed of 5 ms -1. Great work done by the engine is ... A. 1,000 joules B. 2,500 joules C. 5,000 joules D. 12,500 joules E. 25,000 joules (From Ebtanas 1994) Discussion Effort. change in kinetic energy of objects: W = 1/2 m Δ (v 2) W = 1/2 x 1 000 x 5 2 W = 12 500 joules
Note: If you know two or velocity v, then v squared his first deducted, not deductible continues squared!.
Fisikastudycenter.com-Example Problem and Discussion of Heat and
Expansion, High School Grades Matter Physics 10 (X) include specific
heat, latent heat (melting heat, steam heat), heat exchange principle /
Black and expansion length principle, the volume of a material. 