Heat is supplied uniformly at the rate of 100W to 1.0 x 10\(^{-2}\)kg of a liquid for 20 seconds. If the temperature of the liquid rises by 5\(^o\)C, then the specific heat capacity of the liquid is
2.0 x 10\(^2\)Jkg\(^-1\)K\(^-1\)
2.0 x 10\(^2\)Jkg\(^-1\)
4.0 x 10\(^4\)Jkg\(^-1\)K\(^-1\)
4.0 x 10\(^4\)Jkg\(^-1\)
Correct answer is C
Heat energy = Electrical Energy
MC\(\theta\) = power x time
5 x 1 x 10\(^{-2}\) x C = 100 x 20
C = \(\frac{200}{5 \times 10^2}\)
= 4 x 10\(^4\) J/kg/\(^oC\)