The amount of heat that needs to be removed from 100 g of 85°c water to make -5°c ice is 70194 J of heat.
The amount of heat that needs to be removed is calculated below.
Heat changes involved are as follows:
Heat change from water at 85°C to water at 0°C = mc₁ΔT₁
Heat change from water at 0°C to ice at 0°C = mL
Heat change from ice at 0°C to ice at -5°C = mc₂ΔT₂
where
m is mass = 100 g
c₁ is the specific heat capacity of water = 4.184 J/g/K
ΔT₁ is temperature change = 85°C
L is the latent heat of fusion of ice = 336 J/g
c₂ is the specific heat capacity of ice = 2.06 J/g/K
ΔT₁ is temperature change = 5°C
Heat change = (100 * 85 * 4.184) + (100 * 336) + (100 * 5 * 2.06)
Heat change = 70194 J
Learn more about heat changes at: https://brainly.com/question/24298104
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