Question

Water at a temperature of 15°C produces ice cubes at a temperature of −5°C. Water is taken in at the
rate of 1 kg every 7 minutes and the input power to the machine is 300 W. The specific heat capacity of
water is 4187 Jkg−1K

−1, the specific heat capacity of ice is 2100 Jkg−1K

−1 and the specific latent heat of

fusion is 335 Jkg−1
i) Calculate the heat energy extracted per kg of ice produced
ii) Calculate the thermal efficiency of the ice making machine.

Answer 1

Answer on Question #72013, Physics / Molecular Physics | Thermodynamics

Question:

Water at a temperature of $15{}^{\circ}\mathrm{C}$ produces ice cubes at a temperature of $-5{}^{\circ}\mathrm{C}$. Water is taken in at the rate of 1 kg every 7 minutes and the input power to the machine is 300 W.

The specific heat capacity of water is $4187 \frac{J}{kg*K}$

The specific heat capacity of ice is $2100 \frac{J}{kg*K}$

The specific latent heat of fusion is $335 \, kJ/kg$

i) Calculate the heat energy extracted per kg of ice produced

ii) Calculate the thermal efficiency of the ice making machine.

Solution:

i) For producing the ice at a temperature of $-5{}^{\circ}\mathrm{C}$, first it's needed to cool the water to $0{}^{\circ}\mathrm{C}$, then freeze the water into ice, and finally cool the ice:

$$Q = c_{\text{water}} m (T_{15} - T_0) + \lambda m + c_{\text{ice}} m (T_0 - T_{-5})$$ $$Q_m = c_{\text{water}} (T_{15} - T_0) + \lambda + c_{\text{ice}} (T_0 - T_{-5}) - \text{per one kilogram}$$ $$Q_m = 4187 * 15 + 335000 + 2100 * 5 = 408.3 \frac{kJ}{kg}$$

ii) Thermal efficiency: $\alpha = \frac{Q_{\text{input}} - Q_{\text{process}}}{Q_{\text{input}}}$

$$Q_{\text{input}} = 300 * 7 * 60 = 126 \frac{kJ}{kg} < Q_m - \text{this machine cannot produce 1 kg every 7 minutes}$$