Question

A liquid of mass 0.2 kg and temperature 135°C is cooled to 25°C. If the specific heat capacity of the liquid is 750 J kg⁻¹°C⁻¹, find the heat energy given out. [Hint: θ_F = (135 – 25) = 110°C]

Answer 1

Fall in temperature θ_F = (135− 25) = 110°C

∴ Heat = mcθ_F

= 0.2 kg × 750 J kg⁻¹°C⁻¹ × 110°C

= 16500 J