Final answer:
To heat 36.0 grams of ice from -10.5°C to water at 22.5°C, a total of 123145.13 J of energy is needed.
Step-by-step explanation:
To determine the amount of energy needed to heat the ice, we will break down the process into two steps:
First, we need to calculate the energy required to raise the temperature of the ice from -10.5°C to 0°C.
Second, we need to calculate the energy required to melt the ice at 0°C and raise the temperature of the resulting water from 0°C to 22.5°C.
For the first step, we use the specific heat of solid ice, which is 2.108 J/g °C:
Energy = mass × specific heat × temperature change
Energy = 36.0 g × 2.108 J/g °C × (0°C - (-10.5°C))
Energy = 36.0 g × 2.108 J/g °C × 10.5°C
Energy = 794.884 J
For the second step, we need to calculate the energy required to melt the ice and raise the temperature of the resulting water:
Energy = energy to melt the ice + energy to raise the temperature of the water
Energy to melt the ice = mass × heat of fusion
Energy to melt the ice = 36.0 g × 79.9 cal/g × (4.184 J/cal)
Energy to melt the ice = 119910.816 J
Energy to raise the temperature of the water = mass × specific heat of liquid water × temperature change
Energy to raise the temperature of the water = 36.0 g × 4.182 J/g °C × (22.5°C - 0°C)
Energy to raise the temperature of the water = 3439.43 J
Adding the two energies together, we get:
Total energy = 794.884 J + 119910.816 J + 3439.43 J
Total energy = 123145.13 J