How much heat is required to convert solid ice with a mass of 810 g and at a temperature of -18.0 °C to liquid water at a temperature of 62.5 °C? (The specific heat of ice is cice = 2100 J/kgK, the specific heat of water is cwater = 4186.8 J/kgK, and the heat of fusion for water is: Lf = 334 kJ/kg.)
To calculate the amount of heat required to convert solid ice to liquid water, we need to consider the following steps:
Step 1: Calculate the heat required to raise the temperature of the ice from -18.0 °C to 0 °C.
Step 2: Calculate the heat required to melt the ice at its melting point.
Step 3: Calculate the heat required to raise the temperature of the water from 0 °C to 62.5 °C.
Step 4: Sum up the results from steps 1, 2, and 3 to find the total heat required.
Let's go through each step in detail:
Step 1: Calculate the heat required to raise the temperature of the ice from -18.0 °C to 0 °C.
The specific heat capacity of ice (cice) is given as 2100 J/kgK. To calculate the heat required (Q1), we can use the formula:
Q1 = mass * cice * ΔT1
where:
- mass is the mass of ice (810 g = 0.81 kg)
- cice is the specific heat capacity of ice (2100 J/kgK)
- ΔT1 is the temperature change (0 °C - (-18.0 °C) = 18.0 °C)
Plugging in the values, we get:
Q1 = 0.81 kg * 2100 J/kgK * 18.0 °C
= 30618 J
Step 2: Calculate the heat required to melt the ice at its melting point.
The heat of fusion for water (Lf) is given as 334 kJ/kg. To calculate the heat required (Q2), we can use the formula:
Q2 = mass * Lf
where:
- mass is the mass of ice (810 g = 0.81 kg)
- Lf is the heat of fusion for water (334 kJ/kg = 334,000 J/kg)
Plugging in the values, we get:
Q2 = 0.81 kg * 334,000 J/kg
= 270,540 J
Step 3: Calculate the heat required to raise the temperature of the water from 0 °C to 62.5 °C.
The specific heat capacity of water (cwater) is given as 4186.8 J/kgK. To calculate the heat required (Q3), we can use the formula:
Q3 = mass * cwater * ΔT2
where:
- mass is the mass of water (which is equal to the mass of ice since the ice melted)
- cwater is the specific heat capacity of water (4186.8 J/kgK)
- ΔT2 is the temperature change (62.5 °C - 0 °C = 62.5 °C)
Plugging in the values, we get:
Q3 = 0.81 kg * 4186.8 J/kgK * 62.5 °C
= 216,408 J
Step 4: Calculate the total heat required.
To find the total heat required (Qtot), we sum up the results from steps 1, 2, and 3:
Qtot = Q1 + Q2 + Q3
= 30618 J + 270540 J + 216408 J
= 517,566 J
Therefore, the heat required to convert solid ice with a mass of 810 g and at a temperature of -18.0 °C to liquid water at a temperature of 62.5 °C is 517,566 J.