Questions
A liquid cools from 500C to 450C in 5 min and from 450C to 41.50C in the next 5 min. The temperature of the surrounding is
a
270C
b
40.30C
c
23.30C
d
33.30C
detailed solution
Correct option is D
Applying θ1−θ2t=αθ1+θ22−θ0, we get (where, α is constant) 50−455=α50+452−θ0 …(i)and 45−41.55=α45+41.52−θ0 …(ii)Solving these two equations, we getθ0 = temperature of surrounding = 33.30C.Talk to our academic expert!
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