Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.270 kg and a temperature of 65.7 ∘ C .

How much ice at a temperature of -14.5 ∘C must be dropped into the water so that the final temperature of the system will be 28.0 ∘C ?

Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×10⁵J/kg .

Answer 1

The water cools from 65.7° to 28° Celsius. That is a change of 37.7°C. This is the same scale as Kelvin so the change is 37.7 K

Multiply (0.270 kg)(37.7 K)(4190 J/(kg⋅K)) = 42,650 Joules lost by the water. This same amount of energy is gained by the ice.

The final temperature of the ice is 28°C, and we have a certain mass (M) we are to find: Take M*(28 K)(4190 J/(kg⋅K)) is the amount of energy from freezing to 28°C.

Then M*(3.34 x 10^5 J/kg) is the energy during melting. Then M*(14.5 K)(2100 J/(kg⋅K)) is the energy from -14.5°C to 0°C.

Add those three expressions, set equal to 42,650 Joules and solve for M.

M*(28 K)(4190 J/(kg⋅K)) + M*(3.34 x 10^5 J/kg) + M*(14.5 K)(2100 J/(kg⋅K)) = 42,650

=> M(481770) = 42650

=> M = 0.0885 kg