A metal of mass 200g at a temperature of 100oC is placed in 100g of water at 25oC in a container of negligible…

Analyze the question A hot metal is dropped into cooler water. Heat lost by the metal equals heat gained by the water, since the container's heat capacity is negligible.

Concept Heat transferred: Q = mcΔθ Principle of heat exchange: heat lost by hot body = heat gained by cold body

Solution

Metal:

• m = 200 g = 0.2 kg
• Δθ = 100 − 30 = 70 K
• c = unknown

Water:

• m = 100 g = 0.1 kg
• c = 4200 J kg⁻¹ K⁻¹
• Δθ = 30 − 25 = 5 K

Heat gained by water: Q = 0.1 × 4200 × 5 = 2100 J

Heat lost by metal: Q = 0.2 × c × 70 = 14c

Equating the two: 14c = 2100 c = 2100 ÷ 14 = 150 J kg⁻¹ K⁻¹

Answer The specific heat capacity of the metal is 150 J kg⁻¹ K⁻¹, which is option A.

Tip Always convert grams to kilograms before substituting when the specific heat capacity is given in J kg⁻¹ K⁻¹. In mixture problems, find each temperature change as (higher − lower) for the cooling body and (final − initial) for the warming body, so both are positive. Then set heat lost = heat gained.