Final answer:
The heat transfer required to raise the temperature of water from 0°C to 30.0°C is 10.8 kJ. To first melt 0.800 kg of ice and then raise its temperature, the heat transfer required is 267.2 kJ + 11.2 kJ.
Step-by-step explanation:
(a) To calculate the heat transfer required to raise the temperature of 0.800 kg of water from 0°C to 30.0°C, we can use the equation:
Q = mcΔT
where Q is the heat transfer, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The specific heat capacity of water is 4.184 J/g°C.
Calculating the heat transfer:
Q = (0.800 kg)(4.184 J/g°C)(30.0°C - 0°C)
Q = 10.8 kJ
(b) To first melt 0.800 kg of 0°C ice and then raise its temperature, we need to consider the heat transfer required for both processes. The heat transfer required to melt the ice can be calculated using:
Q = mL
where Q is the heat transfer, m is the mass of the ice, and L is the latent heat of fusion of ice which is 334 J/g.
Calculating the heat transfer:
Q = (0.800 kg)(334 J/g)
Q = 267.2 kJ
Then, to raise the temperature of the melted ice from 0°C to 30.0°C, we can use the same equation as in part (a):
Q = mcΔT
Calculating the heat transfer:
Q = (0.800 kg)(4.184 J/g°C)(30.0°C - 0°C)
Q = 11.2 kJ
(c) The answer supports the contention that ice is more effective in absorbing energy because it requires a higher heat transfer to first melt the ice and then raise its temperature compared to just raising the temperature of water.