Question

In order to cool 1 ton of water at \(20^{\circ} \mathrm{C}\) in an insulated tank, a person pours \(80 \mathrm{kg}\) of ice at \(-5^{\circ} \mathrm{C}\) into the water. Determine the final equilibrium temperature in the tank. The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg},\) respectively.

Short answer

Answer: The final equilibrium temperature is approximately \(13.428^{\circ}\mathrm{C}\).

Step-by-step solution

Calculate heat gained by ice to reach melting point

Note that the phase change of ice happens at \(0^{\circ}\mathrm{C}\), so the first step for heat calculation is to warm the ice to that temperature. $$Q_{gain1} = m_{ice} \cdot C_{ice} \cdot \Delta T = 80\,\mathrm{kg} \cdot 2.093\,\mathrm{kJ/kg^{\circ}C} \cdot (0 - (-5))^{\circ}\mathrm{C} = 837.2\,\mathrm{kJ}$$

Calculate heat needed for ice to change phase to water

Next, we need to calculate the heat gained by the ice during the phase change from solid to liquid: $$Q_{gain2} = m_{ice} \cdot H_{fusion} = 80\,\mathrm{kg} \cdot 333.7\,\mathrm{kJ/kg} = 26,696\,\mathrm{kJ}$$

Add heat gained to reach melting point and heat for phase change

The total heat gained by the ice is the sum of heat gained to reach the melting point and the heat gained for the phase change: $$Q_{total\_gain} = Q_{gain1} + Q_{gain2} = 837.2\,\mathrm{kJ} + 26,696\,\mathrm{kJ} = 27,533.2\,\mathrm{kJ}$$

Calculate heat loss by water to achieve equilibrium temperature

Let's denote \(T_{eq}\) as the equilibrium temperature. Then we have: $$Q_{loss} = m_{water} \cdot C_{water} \cdot \Delta T = 1000\,\mathrm{kg} \cdot 4.186\,\mathrm{kJ/kg^{\circ}C} \cdot (20 - T_{eq})^{\circ}\mathrm{C}$$

Set heat gained equal to heat loss and solve for equilibrium temperature

Now, we know that \(Q_{total\_gain} = Q_{loss}\), so we can solve for the equilibrium temperature by equating their expressions: $$27,533.2\,\mathrm{kJ} = 1000\,\mathrm{kg} \cdot 4.186\,\mathrm{kJ/kg^{\circ}C} \cdot (20 - T_{eq})^{\circ}\mathrm{C}$$ Dividing both sides by \(1000\,\mathrm{kg} \cdot 4.186\,\mathrm{kJ/kg^{\circ}C}\), we get: $$6.572^{\circ}\mathrm{C} = 20 - T_{eq}^{\circ}\mathrm{C}$$ This leads us to the final equilibrium temperature: $$T_{eq} = 20 - 6.572 = 13.428^{\circ}\mathrm{C}$$ In conclusion, the final equilibrium temperature in the tank is approximately \(13.428^{\circ}\mathrm{C}\).