Water of mass 150g at 60\(^o\)c is added to 300g of water at 20\(^o\)c and the mixture is well stirred. Calculate the temperature of the mixture.(neglect heat losses to the surroundings)
A. 33\(^o\)c B. 40\(^o\)c C. 25\(^o\)c D. 10\(^o\)c
Correct Answer: A
Explanation
Water 1 => M = 150g = 0.15kg,  θ\(_2\) = 60\(^o\)C; θ\(_1\) = θ  Q\(_1\) = MC (θ\(_2\) – θ\(_1\))  = 0.15C (60 - θ)  Water 2 => M = 300g = 0.3kg  θ1 = 20\(^o\)C, θ\(_2\) = θ  Q2 = MC (θ\(_2\) – θ\(_1\))  = 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  0.45 θ = 15  θ = 33.33\(^o\)C
