Question

200 g of hot water at 80°C is added to 300
water at 10°C. Neglecting the heat tak
container, calculate the final temperat
mixture of water. Specific heat capacit
= 4200 J kg-1 K-1.​

Answer 1

Answer:

38°C

Explanation:

Let, the final temperature be T°C

Then,

Heat lost by hot water = 0.2×4000×(80-T)

Geat gained by cold water = 0.3×4000×(T-10)

According to the principle of mixures,

0.2×4000×(80-T) = 0.3×4000×(T-10)

=>16-0.2T = 0.3T-T

=>19 = 0.5T

=> T= 19/0.5

=> T= 38°C

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