To solve this problem, we need to consider several steps of heating and phase changes that the water undergoes.

1. Heating the ice from -30°C to 0°C:
The specific heat capacity of ice is 2.09 J/g°C. So, the heat required (q1) can be calculated using the formula q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
q1 = 1kg * 2.09 kJ/kg°C * 30°C = 62.7 kJ

2. Melting the ice at 0°C to water at 0°C:
The heat of fusion of ice is 334 J/g. So, the heat required (q2) can be calculated using the formula q = mL, where L is the heat of fusion.
q2 = 1kg * 334 kJ/kg = 334 kJ

3. Heating the water from 0°C to 100°C:
The specific heat capacity of water is 4.18 J/g°C. So, the heat required (q3) can be calculated using the formula q = mcΔT.
q3 = 1kg * 4.18 kJ/kg°C * 100°C = 418 kJ

4. Boiling the water at 100°C to steam at 100°C:
The heat of vaporization of water is 2260 J/g. So, the heat required (q4) can be calculated using the formula q = mL.
q4 = 1kg * 2260 kJ/kg = 2260 kJ

5. Heating the steam from 100°C to 150°C:
The specific heat capacity of steam is 2.03 J/g°C. So, the heat required (q5) can be calculated using the formula q = mcΔT.
q5 = 1kg * 2.03 kJ/kg°C * 50°C = 101.5 kJ

Adding all these values together gives the total heat required:
q_total = q1 + q2 + q3 + q4 + q5 = 62.7 kJ + 334 kJ + 418 kJ + 2260 kJ + 101.5 kJ = 3176.2 kJ

To convert this to kilowatt-hours, use the conversion factor 1 kJ = 0.00027778 kWh:
q_total = 3176.2 kJ * 0.00027778 kWh/kJ = 0.882 kWh

B. To find the power developed in horsepower, we first need to convert the time to hours:
t = 10 min = 0.16667 hr

Power (P) is defined as the work done (or energy transferred) per unit time, so P = W/t. In this case, the work done is the heat transferred, so:
P = q_total / t = 0.882 kWh / 0.16667 hr = 5.29 kW

To convert this to horsepower, use the conversion factor 1 kW = 1.34102 hp:
P = 5.29 kW * 1.34102 hp/kW = 7.09 hp

Question

To solve this problem, we need to consider several steps of heating and phase changes that the water undergoes.

1. Heating the ice from -30°C to 0°C: 
The specific heat capacity of ice is 2.09 J/g°C. So, the heat required (q1) can be calculated using the formula q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. 
q1 = 1kg * 2.09 kJ/kg°C * 30°C = 62.7 kJ

2. Melting the ice at 0°C to water at 0°C: 
The heat of fusion of ice is 334 J/g. So, the heat required (q2) can be calculated using the formula q = mL, where L is the heat of fusion. 
q2 = 1kg * 334 kJ/kg = 334 kJ

3. Heating the water from 0°C to 100°C: 
The specific heat capacity of water is 4.18 J/g°C. So, the heat required (q3) can be calculated using the formula q = mcΔT. 
q3 = 1kg * 4.18 kJ/kg°C * 100°C = 418 kJ

4. Boiling the water at 100°C to steam at 100°C: 
The heat of vaporization of water is 2260 J/g. So, the heat required (q4) can be calculated using the formula q = mL. 
q4 = 1kg * 2260 kJ/kg = 2260 kJ

5. Heating the steam from 100°C to 150°C: 
The specific heat capacity of steam is 2.03 J/g°C. So, the heat required (q5) can be calculated using the formula q = mcΔT. 
q5 = 1kg * 2.03 kJ/kg°C * 50°C = 101.5 kJ

Adding all these values together gives the total heat required: 
q_total = q1 + q2 + q3 + q4 + q5 = 62.7 kJ + 334 kJ + 418 kJ + 2260 kJ + 101.5 kJ = 3176.2 kJ

To convert this to kilowatt-hours, use the conversion factor 1 kJ = 0.00027778 kWh: 
q_total = 3176.2 kJ * 0.00027778 kWh/kJ = 0.882 kWh

B. To find the power developed in horsepower, we first need to convert the time to hours: 
t = 10 min = 0.16667 hr

Power (P) is defined as the work done (or energy transferred) per unit time, so P = W/t. In this case, the work done is the heat transferred, so: 
P = q_total / t = 0.882 kWh / 0.16667 hr = 5.29 kW

To convert this to horsepower, use the conversion factor 1 kW = 1.34102 hp: 
P = 5.29 kW * 1.34102 hp/kW = 7.09 hp

Knowee AI · Accepted Answer