Heat supplied in changing temp of a substance?

Question

Judy places 0.150 kg of boiling water in a thermos bottle. How many kg of ice at -12.0 C must Judy add to the thermos so that the equillibrium temperature of the water is 75C?

Answer 1

Ice at -12°C to 0°C
= Grams (g) x 12°C x 2.050J/g/°C = g x 24.6
Ice at 0°C to water at 0°C = g x 334J
Water at 0°C to 75°C = g x 4.184 x 75 = 313.8J
Total gained by Ice = 672 J
Heat lost by water = 150g x 4.184 J/g/°C x 25°C
= 15,690 J
15,690 ÷ 672 = 23.35 grams of ice = 0.0234kg

Answer 2

Boiling water is 100ºC, so the temperature depression is 25ºC. Use the specific heat of water, cp to determine the energy lost E = mw*cpw*∆T

For the ice, it will melt at 0ºC. Use the specific heat of ice to get the energy to raise the ice temp 12ºC. E1 = mi*cpi*12ºC.

Add to that the heat of fusion of ice (E2 = hf*mi), and

the the specific heat of the water from the ice to get the energy used to raise the temp of the water to 75ºc. ( E3 = mi*cpI*75ºC)

Add all these together (E1 + E2 + E3) (each term will contain the mass of ice/melted ice) and set it equal to E above.

Solve for mi