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
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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
Heat lost by L1 = heat gained by L2
cL1 = 2cL2
100 x 2cL2 x (78 - 66)
= xg x cL2 x (66 - 50)
xg = 150g