Help needed...I have seen this question answered by a few people but no matter h
ID: 1321648 • Letter: H
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Help needed...I have seen this question answered by a few people but no matter how it's solved the system is not accepting answers. Does anyone have a new method? I will rate!!! The same amount of heat entering identical masses of different substances produces different temperature changes. Calculate the final temperature when 2.00 kcal of heat enters 1.43 kg of the following, originally at 30.2C. The specific heat capacity for each material is given in square brackets below.
Help needed...I have seen this question answered by a few people but no matter how it's solved the system is not accepting answers. Does anyone have a new method? I will rate!!! The same amount of heat entering identical masses of different substances produces different temperature changes. Calculate the final temperature when 2.00 kcal of heat enters 1.43 kg of the following, originally at 30.2 C. The specific heat capacity for each material is given in square brackets below. (a) water [1.00 kcal/(kg C)] Check your calculations. C (d) mercury [0.0333 kcal/(kg C)] C Check your calculations. C (c) steel [0.108 kcal/(kg C)] C (b) concrete [0.20 kcal/(kg I C)] 30.89Explanation / Answer
Q = mcdT
dT = Q/mc
dT = 2 kcal / 1.43 kg * 0.20 kcal/(kgoC)
dT = 6.993
Tf - Ti = 6.993
Tf = 6.993 + Ti
Tf = 6.993 + 30.2 = 37.193oC
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