A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to 50°C. It will take

50 sec

90 sec

60 sec

48 sec

Solution

48 sec

The rate of cooling α excess of temperature

Newton's of cooling

$\frac{dQ}{dt} = K (T_{2} - T_{1})$

$\frac{80^{0} - 60^{0}}{60^{0}} = K (\frac{80^{0} + 60^{0}}{2} - 30^{0}) \frac{1}{3} = K (4) \frac{60^{0} - 50^{0}}{t} = K (\frac{60^{0} + 50^{0}}{2} - 30^{0}) \frac{10}{t} = K (25) Dividing equation (1) by (2), we get \frac{t}{30} = \frac{40}{25} or t = \frac{40}{25} \times 30 = 48 \sec$

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Mechanical Properties Of Fluids

A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to 50°C. It will take

50 sec

90 sec

60 sec

48 sec

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For Daily Free Study Material Join wiredfaculty

Mechanical Properties Of Fluids

A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to 50°C. It will take 50 sec 90 sec 60 sec 48 sec

Some More Questions From Mechanical Properties of Fluids Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to 50°C. It will take

50 sec

90 sec

60 sec

48 sec