OMG this problem is making me mad

[BuBbLeS01]

OMG this problem is making me mad please help!

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

 

[BuBbLeS01]

nevermind I got it...I had to use 4186.8 instead of 4190!

 

[ogb p]

Hello, I understand that this problem may be causing frustration. However, it is important to approach it with a calm and analytical mindset in order to find the correct solution. Based on the given information, the total amount of heat required can be calculated by breaking the process into three steps: first, heating the ice from -11.5°C to 0°C, then melting the ice at 0°C, and finally heating the liquid water from 0°C to 60°C. By using the specific heat of ice, the heat of fusion for water, and the specific heat of liquid water, the total amount of heat required can be calculated to be 906,000 joules. I hope this helps.