A. 400 J kg-1K-1
B. 500 J kg-1K-1
C. 390 J kg-1K-1
D. 130 J kg-1K-1
Correct Answer:
Option B – 500 J kg-1K-1
Explanation
Heat supplied by the heater in 10 mins
= power x time
= 50 x (10 x 60)
= 30000 J
This heat is equal = MCΔt
= 5 x c x 12
= 60cJ
= 60C = 30,000
C = 30,000/60
C = 500 J kg-1K-1

![An immersion heater rated 400W, 220V is used to heat a liquid of mass 0.5kg. If the temperature of the liquid increases uniformly at the rate of 2.5ºC per second, calculate the specific heat capacity of the liquid. [Assume no heat is lost] An immersion heater rated 400W, 220V is used to heat a liquid of mass 0.5kg. If the temperature of the liquid increases uniformly at the rate of 2.5ºC per second, calculate the specific heat capacity of the liquid. [Assume no heat is lost]](https://erudites.ng/wp-content/plugins/contextual-related-posts/default.png)