Final answer:
a) To raise the temperature of 0.800 kg of water from 0°C to 30.0°C, 10,548 J of heat transfer is necessary.
b) To first melt 0.800 kg of 0°C ice and then raise its temperature, a total heat transfer of 277,229.6 J is required.
A bag containing ice is more effective in absorbing energy than one containing water due to the high latent heat of fusion of ice.
Step-by-step explanation:
To answer question (a), we can use the formula for heat transfer:
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. Plugging in the values, we get:
Q = (0.800 kg)(4186 J/kg·°C)(30.0 °C - 0 °C)
Solving this equation, we find that Q = 10,548 J of heat transfer are necessary.
To answer question (b), we need to consider the heat transfer required to melt the ice and then raise its temperature. First, we calculate the heat transfer needed to melt the ice using the formula:
Q = mL
where Q is the heat transfer, m is the mass of the ice, and L is the latent heat of fusion. Substituting the values, we get:
Q = (0.800 kg)(334,000 J/kg)
This gives us a heat transfer of Q = 267,200 J required to melt the ice. Next, we calculate the heat transfer needed to raise the temperature of the water using the formula:
Q = mcΔT
Substituting the values, we get:
Q = (0.800 kg)(4186 J/kg·°C)(30.0 °C)
This gives us a heat transfer of Q = 10,029.6 J required to raise the temperature of the water from 0 °C to 30.0 °C. Adding these two heat transfers together, we find the total heat transfer required is:
Total Q = 267,200 J + 10,029.6 J = 277,229.6 J
Therefore, to first melt the ice and then raise its temperature, 277,229.6 J of heat transfer is required.