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
Verified Correct Answer: A.
Physical Principles & Numerical Solution
Water 1 => M = 150g = 0.15kg,
\theta\(_2\) = 60\(^o\)C; \theta\(_1\) = \theta
Q\(_1\) = MC (\theta\(_2\) – \theta\(_1\))
= 0.15C (60 - \theta)
Water 2 => M = 300g = 0.3kg
\theta1 = 20\(^o\)C, \theta\(_2\) = \theta
Q2 = MC (\theta\(_2\) – \theta\(_1\))
= 0.3C (\theta – 20)
Combining Q1 and Q2
0.15C (60 - \theta) = 0.3C (\theta – 20)
9 – 0.15\theta = 0.3\theta – 6
0.3\theta + 0.15\theta = 9 + 6
0.45 \theta = 15
\theta = 33.33\(^o\)C
Source: Myschool Past Questions Archive (Record #47728)