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
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