A metal of mass 200g at a temperature of 100oC is placed in 100g of water at 25oC in a container of negligible heat capacity. If the final steady temperature is 30oC, calculate the specific heat capacity of the metal. [specific heat capacity of the water is 4200 Jkg-1 K-1]
-
150 Jkg-1K-1
- 300 Jkg-1K-1
- 320 Jkg-1K-1
- 1960 Jkg-1K-1
The correct answer is: A
Explanation
MC\(\theta\)metal = MC\(\theta _{water}\)0.2 x C x 70 = 0.1 x 4200 x 5
C = 150