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: Option 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
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