## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 17 - Work, Heat, and the First Law of Thermodynamics - Exercises and Problems - Page 497: 18

#### Answer

The initial temperature of the water was $73.5^{\circ}C$

#### Work Step by Step

We can find the mass of the water as: $m_w = \rho~V$ $m_w = (1000~kg/m^3)(200~mL)(\frac{1~m^3}{10^6~mL})$ $m_w = 0.200~kg$ The heat energy lost by the water will be equal to the heat energy gained by the glass thermometer. We can find the initial temperature of the water: $m_w~c_w~\Delta T_w = m_g~c_g~\Delta T_g$ $m_w~c_w~(T - 71.2^{\circ}C) = m_g~c_g~\Delta T_g$ $m_w~c_w~T = m_g~c_g~\Delta T_g+m_w~c_w~(71.2^{\circ}C)$ $T = \frac{m_g~c_g~\Delta T_g+m_w~c_w~(71.2^{\circ}C)}{m_w~c_w}$ $T = \frac{(0.0500~kg)(750~J/kg~C^{\circ})(71.2^{\circ}C-20.0^{\circ}C)+(0.200~kg)(4186~J/kg~C^{\circ})(71.2^{\circ}C)}{(0.200~kg)(4186~J/kg~C^{\circ})}$ $T = 73.5~^{\circ}C$ The initial temperature of the water was $73.5^{\circ}C$.

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