A 50 g ice cube at -10°C is added to 200 g of water at 30°C. The final temperature of the mixture is:
(Specific heat of water = 1 cal g\(^{-1}\)°C\(^{-1}\), latent heat of fusion of ice = 80 cal g\(^{-1}\), specific heat of ice = 0.5 cal g\(^{-1}\)°C\(^{-1}\))
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When mixing substances at different temperatures, use the principle of conservation of energy to balance the heat gained and lost. Make sure to account for the latent heat required for phase changes, like melting ice.
We will apply the principle of conservation of energy: the heat lost by the warm water will be equal to the heat gained by the ice. The energy required to bring the system to thermal equilibrium is split between:
The heat required to raise the temperature of the ice to 0°C.
The heat required to melt the ice.
The heat required to raise the temperature of the melted ice (now water) to the final equilibrium temperature. Step 1: Calculate the heat required to raise the temperature of the ice from -10°C to 0°C.
The formula for heat is:
\[
Q = m C \Delta T
\]
Where:
\( m \) is the mass of the ice,
\( C \) is the specific heat of the ice,
\( \Delta T \) is the temperature change.
For the ice:
\[
Q_{\text{ice}} = (50 \, \text{g}) \times (0.5 \, \text{cal/g°C}) \times (0°C - (-10°C)) = 50 \times 0.5 \times 10 = 250 \, \text{cal}
\]
Step 2: Calculate the heat required to melt the ice at 0°C.
The heat required to melt the ice is given by:
\[
Q_{\text{melt}} = m L_f
\]
Where:
\( L_f \) is the latent heat of fusion of ice.
For the ice:
\[
Q_{\text{melt}} = (50 \, \text{g}) \times (80 \, \text{cal/g}) = 4000 \, \text{cal}
\]
Step 3: Calculate the heat required to raise the temperature of the melted ice to the final equilibrium temperature.
Let the final temperature be \( T_f \) (in °C). The heat required to raise the temperature of the melted ice from 0°C to \( T_f \) is:
\[
Q_{\text{melted ice}} = m C_{\text{water}} \Delta T
\]
Where:
\( C_{\text{water}} = 1 \, \text{cal/g°C} \) is the specific heat of water.
For the melted ice:
\[
Q_{\text{melted ice}} = (50 \, \text{g}) \times (1 \, \text{cal/g°C}) \times (T_f - 0°C) = 50 \times (T_f)
\]
Step 4: Calculate the heat lost by the water.
The heat lost by the warm water is:
\[
Q_{\text{water}} = m C_{\text{water}} \Delta T
\]
For the water:
\[
Q_{\text{water}} = (200 \, \text{g}) \times (1 \, \text{cal/g°C}) \times (30°C - T_f)
\]
\[
Q_{\text{water}} = 200 \times (30 - T_f)
\]
Step 5: Apply conservation of energy.
The total heat gained by the ice is equal to the heat lost by the warm water:
\[
Q_{\text{ice}} + Q_{\text{melt}} + Q_{\text{melted ice}} = Q_{\text{water}}
\]
Substitute the values:
\[
250 + 4000 + 50 T_f = 200 (30 - T_f)
\]
Step 6: Solve for \( T_f \).
Simplify the equation:
\[
4250 + 50 T_f = 6000 - 200 T_f
\]
\[
50 T_f + 200 T_f = 6000 - 4250
\]
\[
250 T_f = 1750
\]
\[
T_f = \frac{1750}{250} = 7°C
\]
Thus, the final temperature of the mixture is \( \boxed{7°C} \).
