If 10 kJ of heat are added to a 15.5 g ice cube

Question

at -5 degrees C, what will be the resulting state and temperature (in C) of the water?

liquid, 13.9

liquid, 72

vapor, 103

vapor, 134

solid, 4.85

Answer 1

data courtesy of:

http://www.engineeringtoolbox.com/water-...

for ice:

Latent heat of melting - 334 J/k

Specific heat capacity ice - 2.108 J/gK

to warm the ice to O celsius:

dH = mCdT

(dH = 15.5grams) (2.108 J/gK)(5 Celsius)  = 163.35 joules

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to melt the ice:

15.5grams @ 334 J/K = 5177 joules

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subtotal for ice @-5C --> liquid @ 0 C = 5340.35 Joules

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there were 10,000 Joules - 5340.35 Joules used = 4659.65 J available

to heat liquid water from 0 to ?

dH = m C dT

4659.65 J = (15.5) (4.184J/g-C) (dT)

dT = 71.85

your answer: water @ 72 C  aka "liquid, 72"