Calorimetry Numerical **Question: 10** [from Selina ICSE class 10 Physics book exercise]

**45 g of water at 50 ^{o} C in a beaker is cooled when 50 g of copper at 18^{o}C is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is 0.39 J g-1 K-1 and that of water is 4.2 J g-1 K-1. State the assumptions used.**

**Solution:**

Mass of water (m1) = 45 g

Temperature of water (T1) = 500 C

Mass of copper (m2) = 50 g

Temperature of copper (T2) = 180 C

Final temperature (T) =?

The specific heat capacity of the copper c2 = 0.39 J / g / K

The specific heat capacity of water c1 = 4.2 J / g / K

m1c1 (T1 – T) = m2c2 (T – T2)

T = (m1c1T1 + m2c2T2) / (m2c2 + m1c1)

T =[ (45 × 4.2 × 50) + (50 × 0.39 × 18)] /[ (45 × 4.2) + (50 × 0.39)]

T = (9450 + 351) / (189 + 19.5)

T = 9801 / 208.5

T = 470 C