Question

When water of mass 70 g and temperature 50 °C is added to water of mass 30 g, the maximum temperature of the mixture is found to be 41 °C. Find the temperature of water of mass 30 g before hot water was added to it.​

Answer 1

m1= 70g, t1= 50°C

m2= 30g, t2=?

c= 4.2 J/g°C

Let final temperature be t= 41°C.

By principle of mixtures

Heat lost by hot water= Heat gained by cold water

m1c(t1-t)= m2c(t-t2)

$70 \times 4.2(50 - 41) = 30 \times 4.2(41 - t2)$

$70 \times 4.2 \times 9 = 30 \times 4.2(41 - t2)$

$2646 = 30 \times 4.2(41 - t2)$

$88.2 = 4.2(41 - t2)$

$21 = 41 - t2$

$t2 = 20$

Therefore, temperature of 30g of water is 20°C

Answer 2

Answer:

Thank you very much for the answer

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