Answer to Question #173173 in Mechanics | Relativity for JM.C

Question

What is the final temperature of 8575 cal of heat when removed from 15.0 g steam at 100⁰C?

Answer 1

$\Delta{Q}= 8575 \ cal;\Delta{Q}\approx4.2*8575=36015\ J=36\ kJ$

$m= 15g = 0.015\ kg$

$T = 100\degree +273=373\ K$

$Q_1 \text{ the amount of heat in the transition from steam to liquid}$

$L=2260\frac{kJ}{kg}$

$Q_1=Lm= 2260*0.015=33.9\ kJ$

$Q_2= \Delta{Q}-Q_1=36-33.9= 2.1\ kJ$

$Q_2\text{ the amount of heat when the temperature changes}$

$Q_2= cm\Delta{T}$

$c= 4.2\frac{kJ}{kg*K}$

$\Delta{T}= \frac{Q_2}{cm}=\frac{2.1}{4.2*0.015}=33\ K$

$T_2= T-\Delta{T}=373-33= 340\ K=67\degree$

Answer: $67\degree\text{ final temperature}$