  Question

# Mass m of a liquid A is kept in a cup at a temperature of 90∘C. When placed in a room having temperature of 20∘C, it takes 5 min for the temperature of the liquid to drop to 30∘C. Another liquid B having nearly the same density as A and of mass m, kept in another identical cup at 50∘C takes 5 min for its temperature to fall to 30∘C when placed in a room having temperature 20∘C. If the two liquids at 90∘C and 50∘C 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.72∘C69.7∘C66∘C52∘C

Solution

## The correct option is C 66∘CSame 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(θav−θ0) ⇒(90−30)5=ksA(90+302−20) ⇒12=ksA(40) ⇒sA=10k3 For coolling of liquid B, we have: 50−305=ksB(50+302−20) ⇒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(90−θ)=m.sB(θ−50) ⇒10k3(90−θ)=5k(θ−50) ⇒180−2θ=3θ−150 ⇒5θ=330⇒θ=66∘C  Suggest corrections   