Question

A solid of mass 25 g (sp. heat capacity 0.8 J/g°C) and at 120°C is placed in 100 g of water at 20°C. Calculate the final temperature of the mixture. Specific heat capacity of water is 4.2 Jg⁻¹°C⁻¹.

Answer 1

Substance	Mass	S.H.C.	Initial Temp.	Final Temp. = x
Hot solid	25 g	0.8 Jg−1°C−1	120°C	θF = (120°C − x)
Cold water	100 g	4.2 Jg−1°C−1	20°C	θR = (x − 20°C)

Heat given out by the solid = mcθ_F

= 25 × 0.8 × (120 − x)

= 2400 − 20 x

Heat absorbed by cold water = mcθ_R

= 100 × 4.2 × (x − 20)

= 420 x − 8400

Heat given out by the solid = Heat absorbed by water

2400 − 20 x = 420 x − 8400

⇒ 440 x = 10800

∴ x = 24.54°C