To calculate the amount of heat absorbed by ice to become water, we need to consider the heat required for each phase change.
The total heat absorbed can be calculated using the following steps:
Heat absorbed to raise the temperature of the ice from -20.0°C to 0.0°C: The specific heat capacity of ice is approximately 2.09 J/g°C. The mass of the ice is 1 * 10^2 g. The temperature change is 0.0°C - (-20.0°C) = 20.0°C.
So, the heat absorbed for this temperature change is: Q1 = mass * specific heat capacity * temperature change = 1 * 10^2 g * 2.09 J/g°C * 20.0°C = 4,180 J
Heat absorbed for the phase change from ice at 0.0°C to water at 0.0°C: The heat of fusion (also known as the latent heat of fusion) for ice is approximately 334 J/g.
So, the heat absorbed for this phase change is: Q2 = mass * heat of fusion = 1 * 10^2 g * 334 J/g = 33,400 J
The total heat absorbed by the ice is the sum of Q1 and Q2: Total heat absorbed = Q1 + Q2 = 4,180 J + 33,400 J = 37,580 J
Therefore, 1 * 10^2 grams of ice at -20.0°C requires 37,580 joules of heat to become water at 0.0°C.