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

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