Answer:
70 °C
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
