## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 14 - Problems - Page 535: 40

#### Answer

$243~grams~$ of ice should be added to the tea. The answer for Problem 39 increases by 35%

#### Work Step by Step

We can find the mass of the tea: $m_t = (1000~kg/m^3)(2.00\times 10^{-4}~m^3) = 0.20~kg$ The heat required to make ice increase in temperature , melt, and increase in temperature as water is equal in magnitude to the loss of heat from the tea and the glass as they cool: $m_i~c_i~\Delta T_i+m_i~L+m_i~c_w~\Delta T_w = m_t~c_w~\Delta T+m_g~c_g~\Delta T$ $m_i~(c_i~\Delta T_i+L+c_w~\Delta T_w) = m_t~c_w~\Delta T+m_g~c_g~\Delta T$ $m_i = \frac{m_t~c_w~\Delta T+m_g~c_g~\Delta T}{c_i~\Delta T_i+L+c_w~\Delta T_w}$ $m_i = \frac{(0.20~kg)(4186~J/kg~C^{\circ})(85.0~C^{\circ})+(0.350~kg)(837~J/kg~C^{\circ})(85.0~C^{\circ})}{(2100~J/kg~C^{\circ})(10.0~C^{\circ})+(333\times 10^3~J/kg)+(4186~J/kg~C^{\circ})(10.0~C^{\circ})}$ $m_i = 0.243~kg$ $243~grams~$ of ice should be added to the tea. We can find percentage that the answer for Problem 39 increases: $\frac{243~g-180~g}{180~g}\times 100\% = 35\%$ The answer for Problem 39 increases by 35%.

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