How much heat is required to raise 250.0 g of ice at a temperature -15.0°C to 105.0°C?

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To calculate the amount of heat required to raise the temperature of a substance, we use the formula:

Q = mcΔT

Where: Q = heat energy (in joules) m = mass of the substance (in grams) c = specific heat capacity of the substance (in J/g°C) ΔT = change in temperature (in °C)

For ice at -15.0°C to 0°C, we use the specific heat capacity of ice (c_ice = 2.09 J/g°C) For melting ice, we use the heat of fusion of ice (334 J/g) For water at 0°C to 100°C, we use the specific heat capacity of water (c_water = 4.18 J/g°C) For boiling water at 100°C to 105°C, we use the specific heat capacity of water (c_water = 4.18 J/g°C)

Calculations:

1. Heat required to raise the temperature of ice from -15.0°C to 0°C: Q_ice = (250.0 g) * (2.09 J/g°C) * (0°C - (-15.0°C)) = 7837.5 J

2. Heat required to melt the ice at 0°C: Q_melt = (250.0 g) * (334 J/g) = 83500 J

3. Heat required to raise the temperature of water from 0°C to 100°C: Q_water_1 = (250.0 g) * (4.18 J/g°C) * (100°C - 0°C) = 104500 J

4. Heat required to raise the temperature of water from 100°C to 105°C: Q_water_2 = (250.0 g) * (4.18 J/g°C) * (105°C - 100°C) = 522.5 J

Total heat required: Q_total = Q_ice + Q_melt + Q_water_1 + Q_water_2 Q_total = 7837.5 J + 83500 J + 104500 J + 522.5 J Q_total = 195360 J

Therefore, the heat required to raise 250.0 g of ice at -15.0°C to 105.0°C is 195360 joules.