Calculate (in calories) the amount of heat needed to melt 96g of ice at -24 degrees C to water at 28 degrees C.

To calculate the amount of heat needed to melt the ice and increase its temperature, you can use the equation:

Q = m * (ΔT * Cp + ΔHfusion)

Where:
Q is the heat (in calories)
m is the mass of the ice (in grams)
ΔT is the change in temperature
Cp is the specific heat capacity of water (1 calorie/gram °C)
ΔHfusion is the heat of fusion for ice (79.7 calories/gram)

First, let's calculate the amount of heat required to melt the ice:

Q1 = m * ΔHfusion
Q1 = 96g * 79.7 calories/gram
Q1 = 7675.2 calories

Next, let's calculate the amount of heat required to raise the temperature of the resulting water from -24°C to 0°C:

Q2 = m * ΔT * Cp
Q2 = 96g * 24°C * 1 calorie/gram °C
Q2 = 2304 calories

Finally, let's calculate the amount of heat required to raise the temperature of the resulting water from 0°C to 28°C:

Q3 = m * ΔT * Cp
Q3 = 96g * 28°C * 1 calorie/gram °C
Q3 = 2688 calories

To find the total heat, we can add up Q1, Q2, and Q3:

Total heat = Q1 + Q2 + Q3
Total heat = 7675.2 calories + 2304 calories + 2688 calories
Total heat = 12667.2 calories

Therefore, the amount of heat needed to melt 96g of ice at -24 degrees C to water at 28 degrees C is 12667.2 calories.

You need two equations.

1. Within a phase
q = mass x specific heat x (Tfinal-Tinitial). For example, for warming from zero C to 28 C you will have
q = mass x specific heat x (28-0) = ?

2. At the phase change. For example, at the melting point you have
q = mass x heat fusion

So you will have q from -24 to zero, at zero, and from zero to 28. Then add each for the total.