Final answer:
To freeze 1 kg of water at room temperature, 334,400 J of thermal energy must be removed. The electrical energy used to freeze the ice is 83,600 J. The total energy, including waste heat, dumped into the kitchen is 418,000 J. The temperature of the air in the kitchen would rise by approximately 1.001 °C. The efficiency of the freezer is approximately 20%.
Step-by-step explanation:
The amount of thermal energy that must be removed from 1 kg of water at room temperature to freeze it is equal to the latent heat of fusion of the water. The latent heat of water for freezing is 80 cal/gm°C. Since 1 kg of water is equal to 1000 grams, the amount of thermal energy required to freeze it is 80 cal/gm°C * 1000 gm = 80,000 cal. To convert this to Joules, we can use the conversion factor of 1 cal = 4.18 J, so the thermal energy required is 80,000 cal * 4.18 J/cal = 334,400 J.
The electrical energy used to freeze the ice can be calculated using the given information that the freezer needs 1 J of energy for every 3 J of heat it removes. This means that for every 4 J of thermal energy removed, 1 J of electrical energy is used. Since we know that 334,400 J of thermal energy is removed, the electrical energy used is equal to 334,400 J / 4 J = 83,600 J.
The total energy, including waste heat, that is dumped into the kitchen can be calculated by summing up the thermal energy and electrical energy used. So the total energy is 334,400 J + 83,600 J = 418,000 J.
To calculate the temperature rise of the 40 kg of air in the kitchen, we can use the formula Q = mcΔT, where Q is the thermal energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change. The specific heat capacity of air is about 0.2 cal/gm°C, so we can use the conversion factor of 1 cal = 4.18 J to convert to Joules. Substituting the values into the formula, we have Q = (40 kg)(1000 gm/kg)(0.2 cal/gm°C)(4.18 J/cal)(ΔT). Since we know that Q is 418,000 J, we can solve for ΔT as follows: ΔT = 418,000 J / ((40 kg)(1000 gm/kg)(0.2 cal/gm°C)(4.18 J/cal)). Using a calculator, we find that ΔT ≈ 1.001 °C. Therefore, the temperature of the air in the kitchen would rise by approximately 1.001 °C.
The efficiency of the freezer can be calculated using the formula Efficiency = (Useful energy output / Total energy input) * 100%. The useful energy output is the electrical energy used to freeze the ice, which is 83,600 J, and the total energy input is the sum of the thermal energy and electrical energy used, which is 418,000 J. Substituting the values into the formula, we have Efficiency = (83,600 J / 418,000 J) * 100% ≈ 20%.