Document 9-2 PHYSICS- Heat Problems lsheet #1-specific heat capact The energy th
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Document 9-2 PHYSICS- Heat Problems lsheet #1-specific heat capact The energy that transfers from one object to another because of a temperature difference between them is called heat. Units: Heat is given in joules or in calories. ( cal-4.184) Specific heat capacity (abbreviated c) of a substance is the quantity of heat required to warm a gram of that substance by one degree Celsius. Units: Specific heat capacity is given in calg C). By knowing the specific heat capacity of a substance it is possible to determine the amount of heat necessary to cause a specific temperature change in a quantity of that substance. The equation used to do this is the heat formula. Heat Formula: Q mc AT, where Q (in cal) is the energy transferred to or from the substance (that is, the amount of heat), m (in g) is the mass of the substance, AT (in C) is the change in temperature of the substance. Common specific heat capacities: water, c-1 cal (g C) copper, c-1 cal(g C aluminum, c-2 cal(g C; average person, c-83 cal (g C) Note: Most substances have specific heat capacities less than that of water (1) How much heat must be added to 200 g of water to raise its temperature from 20'C to 40C? (2) How much heat must be added to 200 g of water to raise its lemperature from 20C 10 60C? Compare your answer to that of problem #1 . What can you conclude ? (3) How much heat must be added to 200 g ofcopper to raise its temperature from 20 C to 60C ? Conque your answer to that of problem #2, what can you conclude ? Given the above results, which one of the following masses is likcly to reach higher temperatures more quickly if set out in the aflernoon sun: 100 g of water or 100 g of copper? Why? 4) What will be the final temperature of 150 g of 20C water if 3000 cal are added to it ? (5) What will be the final temperature of 150 g of 20'C aluminum if 3000 cal are added to it? (6) A glass containing 225 g of water is placed in the freezer. Ifthe water's temperature changes from 20-C to 2C, then what amount of heat was lost by the water?Explanation / Answer
1.
Q = m*C*dT
m = 200 gm
C = 1 cal/gm-C
dT = 40 - 20 = 20 C
So,
Q = 200*1*20 = 4000 cal
2.
Q = m*C*dT
dT = 60 - 20 = 40 C
Q = 200*1*40 = 8000 cal
from above answer we see that if mass of water remains constant, then to increase water's temperature by T, each time we need fixed energy. In other words energy required to increase temperature by T, remains same for fixed amount of water.
3.
Q = m*C*dT
C = 0.1 cal/gm-C
Q = 200*0.1*40 = 800 cal
Energy require to increse temperature by same amount is different for every material and it depend on specific heat of material.
from Q2 and Q3 we can see that energy required to increase temperature of copper is lower than water, So copper will reach at higher temperature more quickly.
4.
Q = m*C*(Tf - Ti)
Tf = Ti + Q/m*C
Tf = 20 + 3000/(150*1) = 40 C
5.
C for aluminum = 0.2 cal/gm-C
Tf = 20 + 3000/(150*0.2) = 120 C
6.
Q = 225*1*(2 - 20)
Q = -4050 Cal
4050 cal of heat lost by water
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