## College Physics (7th Edition)

Published by Pearson

# Chapter 11 - Heat - Learning Path Questions and Exercises - Exercises - Page 412: 6

6.0 kg

#### Work Step by Step

Given/Known: $Q= 2.0\times10^{6}J$, $\Delta T=(100-20)^{\circ}C=80^{\circ}C$ and $c=4186\,J/(kg\,^{\circ}C)$ Equation to find m: $c=\frac{Q}{m\Delta T}$ or $m=\frac{Q}{c\Delta T}$ Result: $m=\frac{ 2.0\times10^{6}J}{4186\,J/(kg\,^{\circ}C)\times80^{\circ}C}=6.0\,kg$

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