Question

How much energy (in kj) is required to convert 1.34 kg of ice at -16°c to water at 18°c?

Answer 1

The total energy required to convert 1.34 kg of ice at -16°C to water at 18°C is 593.04 kJ, which includes heating the ice up to 0°C, melting the ice, and finally heating the water to 18°C.

Step-by-step explanation:

To calculate the energy required to convert 1.34 kg of ice at -16°C to water at 18°C, we must consider the energy needed for three steps: heating the ice to 0°C, melting the ice at 0°C, and heating the resulting water to 18°C.

• The specific heat capacity of ice is approximately 2.09 kJ/(kg°C). Hence, the energy required to heat 1.34 kg of ice from -16°C to 0°C is: Q1 = (1.34 kg) × (2.09 kJ/(kg°C)) × (16°C) = 44.8384 kJ.
• The latent heat of fusion of ice is 334 kJ/kg. So, the energy required to melt 1.34 kg of ice is: Q2 = (1.34 kg) × (334 kJ/kg) = 447.56 kJ.
• The specific heat capacity of water is approximately 4.18 kJ/(kg°C). Therefore, the energy required to heat 1.34 kg of water from 0°C to 18°C is: Q3 = (1.34 kg) × (4.18 kJ/(kg°C)) × (18°C) = 100.6416 kJ.

Adding the energies from each step, the total energy required is: Qtotal = Q1 + Q2 + Q3 = 44.8384 kJ + 447.56 kJ + 100.6416 kJ = 593.04 kJ.