# A liquid cools from 50^(@)C to 45^(@)C in 5 minutes and from 45^(@)C to 41.5^(@)C in the next 5 minutes. The temperature of the surrounding is

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A liquid cools from 50^(@)C to 45^(@)C in 5 minutes and from 45^(@)C to 41.5^(@)C in the next 5 minutes. The temperature of the surrounding is
A. 27^(@)C
B. 40.3^(@)C
C. 23.3^(@)C
D. 33.3^(@)C

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Applying (theta_(1)-theta_(2))/t = alpha[(theta_(1)+theta_(2))/2 - (theta_(0))] two times, we get where alpha is constant.
(50-45)/5 = alpha[(50+45)/2-(theta_(0))...............(i)
(45-41.5)/(5) = alpha[(45+41.5)/2-(theta_(@))].......................(ii)
Solving these two equations, we get
theta_(@)= temperature of atmosphere = 33.3^(@)C