When 100g of liquids L1 at 78^\circC was mixed with Xg of liquid L2 at 50^\circC, the final temperature was 66^\circC. Given that the specific heat capacity of L2is half that of L1, find x
A
50g
B
100g
C
150g
D
200g
Verified Correct Answer: C.
Physical Principles & Numerical Solution
Heat lost by L1 = heat gained by L2
cL1 = 2cL2
100 x 2cL2 x (78 - 66)
= xg x cL2 x (66 - 50)
xg = 150g
Source: Myschool Past Questions Archive (Record #27907)