JAMB - Physics (1994 - No. 16)
When 100g of liquids L1 at 78°C was mixed with Xg of liquid L2 at 50°C, the final temperature was 66°C. Given that the specific heat capacity of L2is half that of L1, find x
50g
100g
150g
200g
Explication
Heat lost by L1 = heat gained by L2
cL1 = 2cL2
100 x 2cL2 x (78 - 66)
= xg x cL2 x (66 - 50)
xg = 150g
cL1 = 2cL2
100 x 2cL2 x (78 - 66)
= xg x cL2 x (66 - 50)
xg = 150g