A metal cube with a specific heat capacity of .55 J/gºC, a mass of 55grams and a temperature of 85°C
is immersed in water at 20°C. The metal and water
then reach a equilibrium temperature of 23°C. What is
the mass of the water?

148.85 g or 0.14885 kg

Explanation:

Applying,

Heat lost by the metal cube = heat gained by the water.

c'm'(t₁-t₃) = cm(t₃-t₂) Equation 1

Where c' = specific heat capacity of the metal, m' = mass of the metal, c = specific heat capacity of water, m = mass of water, t₁ = Initial temperature of the metal, t₂ = Initial temperature of water, t₃ = equilibrium temperature.

make m the subject of the equation

m = c'm'(t₁-t₃)/c(t₃-t₂) Equation 2

From the questions,

Given: c' = 0.55 J/g.°C, m' = 55 g, t₁ = 85°C, t₂ = 20°C, t₃ = 23°C

Constant: c = 4.2 J/g.°C

Substitute these values into equation 2

m = 0.55×55×(85-23)/(4.2×[23-20])

m = 1875.5/12.6

m = 148.85 g.

m = 0.14885 kg

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