Textbook Solution:

Answer:

To determine

To determine: The final temperature of the mixture.

Answer

Answer: Final temperature of the mixture is $16 ° C$ .

Explanation

Explanation:

Given Info: Mass of ice 75 g, initial temperature of ice is $0 ° C$ , mass of water is 825 g and initial temperature of water is $25 ° C$ .

Part of Heat energy from water melts the ice to water at $0 ° C$ and remaining heat energy raises the temperature of the cold water. Heat lost be warm water is equal to heat gained by cold water.

Formula to calculate the heat gained by cold water is,

$Q_{cold} = m_{ice} L_{f} + m_{ice} c_{water} (T_{f} - T_{i, ice})$

$Q_{cold}$ is the energy gained by cold water from ice,
$m_{ice}$ is the mass of the ice,
$L_{f}$ is the latent heat of ice,
$c_{water}$ is the specific heat of water,
$T_{i, ice}$ is the initial temperature of cold water from ice,
$T_{f}$ is the final temperature of the mixture,

Formula to calculate the heat lost by warm water is,

$Q_{warm} = m_{water} c_{water} (T_{f} - T_{i, water})$

$Q_{warm}$ is the energy lost by warm water,
$m_{water}$ is the mass of warm water,
$T_{i, water}$ is the initial temperature of warm water,

Heat lost by warm water is equal to the Heat gained by cold water from ice.

$Q_{cold} = - Q_{warm}$

Use $m_{ice} L_{f} + m_{ice} c_{water} (T_{f} - T_{i, ice})$ for $Q_{cold}$ and $m_{water} c_{water} (T_{f} - T_{i, water})$ for $Q_{warm}$ to find $T_{f}$ .

$\begin{matrix} m_{ice} L_{f} + m_{ice} c_{water} (T_{f} - T_{i, ice}) = - m_{water} c_{water} (T_{f} - T_{i, water}) \\ m_{ice} L_{f} + m_{ice} c_{water} T_{f} - m_{ice} c_{water} T_{i, ice} = - m_{water} c_{water} T_{f} + m_{water} c_{water} T_{i, water} \\ m_{ice} c_{water} T_{f} + m_{water} c_{water} T_{f} = m_{water} c_{water} T_{i, water} + m_{ice} c_{water} T_{i, ice} - m_{ice} L_{f} \\ (m_{ice} + m_{water}) c_{water} T_{f} = m_{water} c_{water} T_{i, water} + m_{ice} c_{water} T_{i, ice} - m_{ice} L_{f} \end{matrix}$

Substitute 825 g for $m_{water}$ , $4186 J / kg \cdot °C$ for $c_{water}$ , $25 ° C$ for $T_{i, water}$ , 75 g for $m_{ice}$ , $3.33 \times 10^{5} J / kg$ for $L_{f}$ , and 0 for $T_{i, ice}$ to find $T_{f}$ .

$\begin{matrix} (825 g + 75 g) (\frac{1 kg}{10^{3} g}) (4186 J / kg \cdot °C) T_{f} = {\begin{cases} [(825 g) (\frac{1 kg}{10^{3} g}) (4186 J / kg \cdot °C) (25 ° C)] \\ + [(75 g) (\frac{1 kg}{10^{3} g}) (4186 J / kg \cdot °C) (0 ° C)] \\ - [(75 g) (\frac{1 kg}{10^{3} g}) 3.33 \times 10^{5} J / kg] \end{cases}} \\ 3767.4 J / °C T_{f} = 61361.25 J \\ T_{f} = \frac{61361.25 J}{3767.4 J / °C} \\ \approx 16 ° C \end{matrix}$

Conclusion:

The final temperature of the mixture is $16 ° C$ .

Answer:

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