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OMG this problem is making me mad please help!
How much heat is required to convert 2.50 kg of solid ice from a temperature of -11.5°C to liquid water at a temperature of 60.0 °C? (The specific heat of ice is cice = 0.50 kcal/kg°C and the heat of fusion for water is Lf = 79.7 kcal/kg.)
M = 2.5kg
Steps: -11.5 deg C to 0 deg C
Heat fusion at 0 deg C
0 deg C to 60 deg C
I converted bother specific heats to Jewels by multiplying by 1000 and 4.19
0.5 = 2095
79.7 - 333943
Q1 = M*Ci*CH T = 2.5 * 2095 * (0+11.5) = 60231.25J
Q2 = MLf = 2.5 * 333943 = 834857.5J
Q3 = M*Cw*CH T = 2.5 * 4190 * (60-0) = 10475J
Q1 + Q2 + Q3 = 9.06E^5 j
Homework Statement
How much heat is required to convert 2.50 kg of solid ice from a temperature of -11.5°C to liquid water at a temperature of 60.0 °C? (The specific heat of ice is cice = 0.50 kcal/kg°C and the heat of fusion for water is Lf = 79.7 kcal/kg.)
Homework Equations
The Attempt at a Solution
M = 2.5kg
Steps: -11.5 deg C to 0 deg C
Heat fusion at 0 deg C
0 deg C to 60 deg C
I converted bother specific heats to Jewels by multiplying by 1000 and 4.19
0.5 = 2095
79.7 - 333943
Q1 = M*Ci*CH T = 2.5 * 2095 * (0+11.5) = 60231.25J
Q2 = MLf = 2.5 * 333943 = 834857.5J
Q3 = M*Cw*CH T = 2.5 * 4190 * (60-0) = 10475J
Q1 + Q2 + Q3 = 9.06E^5 j
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