Question

80 g of water at 60°C is poured into a vessel containing 100 g of water at 20°C. The final temperature recorded is 30°C. Calculate the thermal capacity of the vessel.

Answer 1

Substance	Mass	S.H.C.	Initial Temp.	Final Temp. = 30°C
Hot water	80 g	4.2 Jg−1°C	60°C	θF = (60 − 30) = 30°C
Cold water	100 g	4.2 Jg−1°C	20°C	θR = (30 − 20) = 10°C
Vessel	?(m)	?(c)	20°C

Let the thermal capacity of the vessel = x = mс

Heat lost by hot water = mcθ_F

= 80 × 4.2 × 30

= 10080 J

Heat gained by cold water = mcθ_R

= 100 × 4.2 × 10

= 4200 J

Heat gained by the vessel = mcθ_R = x × 10°C

Heat gained = Heat lost

4200 J + x × 10°C = 10080 J

\[x =\frac{(10080-4200)}{10} \mathrm{~J}^{\circ} \mathrm{C}^{-1}\]

∴ x = 588 J°C⁻¹