Question

If 100g of water at 40C temperature and 150g.of water 80degree c temperature are mixed, what is final temperature

Answer 1

Given 
Mass of water at 80°C (m1) = 150 g
Temperature (t1) = 80° C
The specific heat capacity of water (c1 ) = 4.2 J/g °C

Mass of water at 40° C (m2) = 100 g
Temperature (t2) = 40° C
the specific heat capacity of water (c2) = 4.2 J/g °C

According to the principle of calorimetry 
Heat energy lost by the hot body = Heat energy gained by the cold body 

m1 c1 (t1 - t) = m2 c2 (t - t2)
150 * 4.2 (80 - t) = 100 * 4.2 (t - 40)
630 (80 - t) = 420 (t - 40)
50400 - 630t = 420t - 16800
50400 + 16800 = 420t + 630t
67200 = 1050t
t = 67200/1050 = 64 °C

The final temperature is 64 °C