Answer:
148.85 g or 0.14885 kg
Step-by-step 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