Water of mass 150g at 60°c is added to 300g of water at 20°c and the mixture is well stirred. Calculate the temperature of the mixture. (neglect heat losses to the surroundings)

A. 33°C
B. 40°C
C. 25°C
D. 10°C

Correct Answer:

Option A – 33°C

Explanation

Water 1 => M = 150g = 0.15kg,

θ2 = 60°C; θ1 = θ

Q1 = MC (θ2 – θ1)

= 0.15C (60 – θ)

Water 2 => M = 300g = 0.3kg

θ1 = 20°C, θ2 = θ

Q2 = MC (θ2 – θ1)

= 0.3C (θ – 20)

Combining Q1 and Q2

0.15C (60 – θ) = 0.3C (θ – 20)

9 – 0.15θ = 0.3θ – 6

0.3θ + 0.15θ = 9 + 6

0.45θ = 15

θ = 33.33°C