Answer:
It would take approximately 81.65 kJ (or 81,650 J) of heat to convert 230.0g of ice at -10C to water at 0C.
Step-by-step explanation:
To calculate the amount of heat needed to convert ice at -10°C to water at 0°C, we need to consider two steps:
1. Heating the ice from -10°C to 0°C (heat required to raise the temperature of ice)
2. Melting the ice into water at 0°C (heat required to change the state of ice)
Let's first calculate the heat required for step 1:
Q1 = m × c × ΔT
where Q1 is the heat required, m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature.
The specific heat capacity of ice is 2.09 J/g°C, and ΔT is (0°C - (-10°C)) = 10°C.
So, Q1 = 230.0 g × 2.09 J/g°C × 10°C = 4827 J
Now, let's calculate the heat required for step 2:
Q2 = m × Lf
where Q2 is the heat required, m is the mass of the ice, and Lf is the latent heat of fusion of ice.
The latent heat of fusion of ice is 334 J/g.
So, Q2 = 230.0 g × 334 J/g = 76820 J
Therefore, the total amount of heat needed to convert 230.0g of ice at -10C to water at 0C is:
Qtotal = Q1 + Q2 = 4827 J + 76820 J = 81647 J
Therefore, it would take approximately 81.65 kJ (or 81,650 J) of heat to convert 230.0g of ice at -10C to water at 0C.