To calculate the amount of heat needed to convert ice at -24 degrees Celsius to water at 28 degrees Celsius, we need to consider the following steps:
Heat absorbed to raise the temperature of ice from -24 degrees Celsius to 0 degrees Celsius.
Heat absorbed to melt the ice at 0 degrees Celsius.
Heat absorbed to raise the temperature of water from 0 degrees Celsius to 28 degrees Celsius.
Step 1: Heat absorbed to raise the temperature of ice:
Q1 = mass × specific heat capacity × change in temperature
= 96 g × 2.09 J/g°C × (0 - (-24))°C
= 96 g × 2.09 J/g°C × 24°C
= 4,777.92 J
Step 2: Heat absorbed to melt the ice:
Q2 = mass × heat of fusion
= 96 g × 333.5 J/g
= 31,968 J
Step 3: Heat absorbed to raise the temperature of water:
Q3 = mass × specific heat capacity × change in temperature
= 96 g × 4.18 J/g°C × (28 - 0)°C
= 11,353.6 J
Total heat needed = Q1 + Q2 + Q3
= 4,777.92 J + 31,968 J + 11,353.6 J
= 48,099.52 J
Therefore, the amount of heat needed to convert 96g of ice at -24 degrees Celsius to water at 28 degrees Celsius is 48,099.52 Joules (J).