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
-
A.
50g -
B.
100g -
C.
150g -
D.
200g
Correct Answer: Option C
Explanation
Heat lost by L1 = heat gained by L2
cL1 = 2cL2
100 x 2cL2 x (78 – 66)
= xg x cL2 x (66 – 50)
xg = 150g