## College Physics (7th Edition)

Published by Pearson

# Chapter 11 - Heat - Learning Path Questions and Exercises - Exercises - Page 413: 14

#### Answer

$T=82.2 C^{\circ}$

#### Work Step by Step

We know that the heat gained by the glass cup is given as $q_{glass}=m_{glass}.C_{glass}.\Delta T_{glass}$ $\implies q_{glass}=(250)(0.840)(T-20)$ $\implies q_{glass}=(210)(T-20)$ Similarly, the heat released by the hot water is given as $q_{H_2O}=m_{H_2O}.C_{H_2O}.\Delta T_{H_2O}$ $\implies q_{H_2O}=(400)(4.184)(90-T)$ $q_{H_2O}=(1673.6)(90-T)$ We know that the heat released by the hot water is equal to the heat gained by the glass that is $(210)(T-20)=(1673.6)(90-T)$ This simplifies to: $T=82.2 C^{\circ}$

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