How many grams of ice at - 13 deg C must be added to 711 grams of water that is initially at a temperature of 87 deg C to produce water at a final temperature of 10 deg C? Assume that no heat is lost to the surroundings and that the container has negligible mass.

The answer to your question is m = 4.7 kg

Explanation:

Data

Ice Water

mass = ? mass = 711 g

T₁ = - 13°C T₁ = 87°C

T₂ = 10°C T₂ = 10°C

Ch = 2090 J/kg°K Cw = 4180 J/kg°K

Process

1. - Convert temperature to kelvin

T₁ = 273 + (-13) = 260°K

T₁ water = 87 + 273 = 360 °K

T₂ = 10 + 273 = 283°K

2. - Write the equation of interchange of heat

- Heat lost = Heat absorbed

- mwCw (T₂ - T₁) = miCi (T₂ - T₁)

-Substitution

- 0.711 (4180) (10 - 87) = m (2090) (10 - (-13))

- Simplification

228842.46 = 48070m

m = 228842.46/48070

-Result

m = 4.7 kg