Question

Some heat is given to 120 g of water and its temp. rises by 10 K. When the same amount of heat is given to 60 g of oil, its temp. rises by 40 K. The Sp. heat of water is 4200 J/kgK. Calculate:
(a) The amount of heat in joules given to water,
(b) The specific heat capacity of the oil.

When 120 g of water receives heat, the temperature rises by 10 K. The temperature of 60 g of oil increases by 40 K when the same amount of heat energy is applied to it. 4200 J kg⁻¹ K⁻¹ is the specific heat capacity of water. Determine the specific heat capacity of oil and the amount of heat energy delivered to water in joules.

Answer 1

For water:

m = 120 g = 0.12 kg

ΔT = 10 K

C = 4200 J/ KgK

Q = m × c × ΔT

= 0.12 × 4200 × 10

= 5040 J

For oil:

Q = 5040 J 

m = 60 g = 0.06 kg

ΔT = 40 K

C = `"Q"/("m" xx Δ"T")`

= `5040/(0.06 xx 40)`

= 2100 J/kg K