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]
- 1100Jkg-1K-1
-
320Jkg-1K-1
- 200Jkg-1K-1
- 176Jkg-1K-1
The correct answer is: B
Explanation
pt = mc\(\theta\)C = \(\frac{pt}{m \theta}\)
= \(\frac{400 \times 1}{0.5 \times 2.5}\)
= 320Jkg-1K-1