Question

200g of water at 90degree Celsius is mixed with 100g of water at 30 degree Celsius. What is the final temperature (C of water =4.2jkg-1 k-1​

Answer 1

70 °C

Step-by-step explanation:

Since heat lost by 200 g of water = heat gained by 100 g of water

-mc(T - T') = m'c(T - T")

where m = 200 g,T' = 90°C, m' = 100 g,T" = 30°C and T = final temperature of mixture, c = specific heat capacity of water = 4.2jkg-1 k-1​.

Substituting the values of the variables into the equation, we have

-mc(T - T') = m'c(T - T")

-200 g(T - 90 °C) = 100 g(T - 30 °C)

(T - 90 °C) = 100 g(T - 30 °C)/-200 g

(T - 90 °C) = -0.5(T - 30 °C)

T - 90 °C = -0.5T + 15 °C

collecting like terms, we have

T + 0.5T = 90 °C + 15 °C

1.5T = 105 °C

T = 105 °C/1,5

T = 70 °C