A solid of mass 50 g at 150°C is placed in 100 g of water at 11°C, when the final temperature recorded is 20°C. Find the specific heat capacity of the solidfSpecific heat capacity of water – 4.2 J g -1 °C -1 )

Given: solid, mass 50 g, at 150 °C; water, mass 100 g, at 11 °C

final temperature 20 °C; c_w = 4.2 J g⁻¹ °C⁻¹ Let the specific heat capacity of the solid be c.

The solid COOLS from 150 °C to 20 °C, so Δθ = 130 °C. The water WARMS from 11 °C to 20 °C, so Δθ = 9 °C.

Heat lost by the solid = 50 × c × 130 = 6 500c Heat gained by the water = 100 × 4.2 × 9 = 3 780 J

Heat lost = heat gained: 6 500c = 3 780 c = 3 780 ÷ 6 500 c = 0.5815

Specific heat capacity of the solid ≈ 0.58 J g⁻¹ °C⁻¹

Each body has its OWN temperature change, and the two are different numbers. The solid falls 130 °C; the water rises only 9 °C. Using the same Δθ for both is the mistake that wrecks this question.

Get each Δθ by subtracting in the direction that makes it positive: for the hot body it is (initial − final), for the cold body (final − initial). Then heat lost = heat gained, and the unknown falls out.

Sanity check: the answer is much smaller than 4.2, which is right — almost every solid has a far lower specific heat capacity than water. An answer bigger than 4.2 means you have the equation inverted.