A metal piece of mass 420 g present at 80°C is dropped in 80 g of water present at 20°C in a calorimeter of mass 84 g. If the final temperature of the mixture is 30°C, then calculate the specific heat capacity of the metal piece. [Specific heat capacity of water = 4.2 Jg-1 °C-1, Specific heat capacity of the calorimeter = 200 Jkg-1 °C-1]

Given: metal 420 g at 80 °C; water 80 g at 20 °C; calorimeter 84 g

final temperature 30 °C c_w = 4.2 J g⁻¹ °C⁻¹, c_cal = 200 J kg⁻¹ °C⁻¹ = 0.2 J g⁻¹ °C⁻¹ Let the specific heat capacity of the metal be S.

The metal COOLS from 80 °C to 30 °C, so Δθ = 50 °C. The water and calorimeter WARM from 20 °C to 30 °C, so Δθ = 10 °C.

HEAT LOST by the metal: = 420 × S × 50 = 21 000S

HEAT GAINED: water = 80 × 4.2 × 10 = 3 360 J calorimeter = 84 × 0.2 × 10 = 168 J total = 3 360 + 168 = 3 528 J

Heat lost = heat gained: 21 000S = 3 528 S = 3 528 ÷ 21 000 S = 0.168

Specific heat capacity of the metal = 0.168 J g⁻¹ °C⁻¹

Watch the calorimeter's units. It is given as 200 J kg⁻¹ °C⁻¹ while the water is in J g⁻¹ °C⁻¹ — two different units in one question. Convert the calorimeter to 0.2 J g⁻¹ °C⁻¹ so everything is per gram, then the masses can all stay in grams.

Using 200 with a mass of 84 g gives 168 000 J for the calorimeter, which dwarfs everything else and produces an absurd answer. Whenever two constants in the same question carry different units, convert one before you start.

The metal falls 50 °C while the water and calorimeter rise 10 °C — separate values of Δθ, as always. And 0.168 is far below 4.2, which is right for a metal.