Final answer:
To raise the temperature of 0.800 kg of water from 0°C to 30.0°C, 100.32 kJ of heat transfer is required. To melt 0.800 kg of ice and then raise its temperature, a total of 367.52 kJ is needed. This illustrates that ice absorbs more energy due to the heat of fusion.
Step-by-step explanation:
Heat Transfer Calculation for Water and Ice
To answer question (a) regarding how much heat transfer is necessary to raise the temperature of 0.800 kg of water from 0°C to 30.0°C, we use the specific heat capacity formula: Q = mcΔT.
Here, Q is the heat transfer, m is the mass, c is the specific heat capacity of water, which is approximately 4.18 kJ/kg°C, and ΔT is the change in temperature.
Plugging in the values, we get:
Q = (0.800 kg)(4.18 kJ/kg°C)(30.0°C - 0°C)
= 100.32 kJ
For question (b) about how much heat transfer is required to first melt 0.800 kg of 0°C ice and then raise its temperature, we have to consider the heat of fusion.
The heat of fusion for water is about 334 kJ/kg.
So, the heat needed to melt the ice is:
Qₘ = (0.800 kg)(334 kJ/kg)
= 267.2 kJ
After the ice has melted into water at 0°C, we can use the previous specific heat capacity formula to calculate the energy required to raise the temperature to 30.0°C:
Qₙ = (0.800 kg)(4.18 kJ/kg°C)(30.0°C) = 100.32 kJ
Adding both the heat of fusion and the heat capacity gives us the total heat transfer for melting and heating the ice:
Total Heat Transfer = Qₘ + Qₙ = 267.2 kJ + 100.32 kJ
= 367.52 kJ
This substantial difference in energy required to bring ice to the same final temperature as water demonstrates that the ice is indeed more effective in absorbing energy.