Final answer:
The amount of heat required to change 100 g of ice at -10°C to water at 90°C is 73080 J.
Step-by-step explanation:
The amount of heat required to change 100 g of ice at -10°C to water at 90°C can be calculated using the formula:
Q = mcΔT + mL
Where Q is the amount of heat, m is the mass of the substance (100 g in this case), c is the specific heat capacity of the substance (2.06 J/g °C for ice and 4.18 J/g °C for water), ΔT is the change in temperature, and L is the latent heat for the phase change.
Since the ice is initially at -10°C, the first step is to heat the ice to its melting point which is 0°C. So, ΔT = 0°C - (-10°C) = 10°C.
The heat required to heat 100 g of ice from -10°C to 0°C is Q1 = mcΔT = 100 g x 2.06 J/g °C x 10°C = 2060 J.
Next, we need to calculate the heat required to melt the ice, also known as the latent heat. The latent heat for the phase change from ice to water is 334 J/g, so the heat required to melt 100 g of ice is Q2 = mL = 100 g x 334 J/g = 33400 J.
Finally, we need to calculate the heat required to heat the water from 0°C to 90°C. So, ΔT = 90°C - 0°C = 90°C.
The heat required to heat 100 g of water from 0°C to 90°C is Q3 = mcΔT = 100 g x 4.18 J/g °C x 90°C = 37620 J.
Adding up the three quantities of heat, we get the total amount of heat required:
Q = Q1 + Q2 + Q3 = 2060 J + 33400 J + 37620 J = 73080 J.