A 5.00 kg block of metal with c = 650 J/(kg*C) at 80.0 °C comes
in contact with a 1.25 kg glass
block at 20.0 °C. They come to
equilibrium at 63.9 °C. What is the
specific heat of the glass?

953.5 J/kg.°C

Explanation:

From the question,

Heat lost by the metal = heat gained by the glass.

cm(t₁-t₃) = c'm'(t₃-t₂) Equation 1

Where c = specific heat capacity of the metal, m = mass of the metal, c' = specific heat capacity of the glass, m' = mass of the glass, t₁ = initial temperature of  metal, t₂ = initial temperature of glass, t₃ = Equilibrium temperature

Make c' the subject of the equation

c' = cm(t₁-t₃)/m'(t₃-t₂) Equation 2

Given: m = 5 kg, c = 650 J/kg.°C, m' = 1.25 kg, t₁ = 80 °C, t₂ = 20 °C, t₃ = 63.9 °C

Substitute these values into equation 2

c' = 5×650(80-63.9)/1.25(63.9-20)

c' = (5×650×16.1)/(1.25×43.9)

c' = 52325/54.875

c' = 953.5 J/kg.°C

960 J/(kg*C)

Explanation:

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