Answer :
The amount of energy required to melt 5.25 kg of ice at 0°C and then warm the resulting water up to 99°C is approximately 3,820,845 joules.
To determine the amount of energy required to melt 5.25 kg of ice and then warm the resulting water, we need to consider two separate processes: the phase change from solid to liquid (melting) and the increase in temperature of the liquid water.
Melting of ice:
To melt the ice, we need to provide the latent heat of fusion, which is the energy required to change the phase from solid to liquid without changing the temperature. The latent heat of fusion for ice is approximately 334,000 J/kg.
Therefore, the energy required to melt the ice can be calculated as follows:
Energy for melting = mass of ice * latent heat of fusion
= 5.25 kg * 334,000 J/kg
= 1,753,500 J
Warming the water:
Once the ice has melted, we have liquid water at 0°C. To warm this water to 99°C, we need to consider the specific heat capacity of water. The specific heat capacity of water is approximately 4,186 J/(kg·°C).
The energy required to increase the temperature of the water can be calculated using the following equation:
Energy for temperature increase = mass of water * specific heat capacity * change in temperature
First, let's calculate the mass of water. Since the ice has melted, all the ice has become water. Therefore, the mass of water is equal to the mass of the ice, which is 5.25 kg.
Next, let's calculate the energy required to increase the temperature from 0°C to 99°C:
Energy for temperature increase = 5.25 kg * 4,186 J/(kg·°C) * (99°C - 0°C)
= 2,067,345 J
Total energy required:
To find the total energy required, we sum up the energy required for melting and the energy required for temperature increase:
Total energy = Energy for melting + Energy for temperature increase
= 1,753,500 J + 2,067,345 J
= 3,820,845 J
Learn more about latent heat at: brainly.com/question/23976436
#SPJ11
