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Mass m of a liquid A is kept in a cup at a temperature of 90C. When placed in a room having temperature of 20C, it takes 5 min for the temperature of the liquid to drop to 30C. Another liquid B having nearly the same density as A and of mass m, kept in another identical cup at 50C takes 5 min for its temperature to fall to 30C when placed in a room having temperature 20C. If the two liquids at 90C and 50C are mixed in a calorimeter where no heat is allowed to leak, find the final temperature of the mixture. Assume that Newton's law of cooling is applicable for the given temperature ranges.
  1. 72C
  2. 69.7C
  3. 66C
  4. 52C


Solution

The correct option is C 66C
Same density of two liquids means that they will occupy the same volume in the identical cups. Therefore, the surface area through which the heat leaks is same for both.
For cooling of liquid A: 
ΔθΔt=ksA(θavθ0)
(9030)5=ksA(90+30220)
12=ksA(40)
sA=10k3
For coolling of liquid B, we have:
50305=ksB(50+30220)
4=ksB(20)
sB=5k
When the two liquids are mixed, let the final temperature of mixture be θ
Heat lost by A = Heat gained by B
m.sA(90θ)=m.sB(θ50)
10k3(90θ)=5k(θ50)
1802θ=3θ150
5θ=330θ=66C

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