To determine the amount of heat needed to transform 500 grams of ice at -20°C into water at 50°C, we need to consider the different phases and temperature ranges involved. Here's a breakdown of the steps involved and the corresponding heat requirements:
Heating the ice from -20°C to 0°C: The specific heat capacity of ice is approximately 2.09 J/g°C. The temperature change is 0°C - (-20°C) = 20°C. Therefore, the heat required for this step is: Q1 = mass × specific heat × temperature change = 500 g × 2.09 J/g°C × 20°C = 20,900 J
Melting the ice at 0°C: The heat of fusion, also known as the latent heat of fusion, for ice is approximately 334 J/g. The mass of the ice is 500 grams. Therefore, the heat required for this step is: Q2 = mass × heat of fusion = 500 g × 334 J/g = 167,000 J
Heating the water from 0°C to 50°C: The specific heat capacity of water is approximately 4.18 J/g°C. The temperature change is 50°C - 0°C = 50°C. Therefore, the heat required for this step is: Q3 = mass × specific heat × temperature change = 500 g × 4.18 J/g°C × 50°C = 104,500 J
The total heat required is the sum of the individual heats for each step: Total heat = Q1 + Q2 + Q3 = 20,900 J + 167,000 J + 104,500 J = 292,400 J
So, approximately 292,400 Joules of heat are needed to transform 500 grams of ice at -20°C into water at 50°C.