(adsbygoogle = window.adsbygoogle || []).push({}); OMG this problem is making me mad please help!!!

1. The problem statement, all variables and given/known data

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.)

2. Relevant equations

3. 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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# Homework Help: OMG this problem is making me mad please help

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