Question

A liquid initially at 70°C cools to 55°C inp 5 minutes
and 45°C in 10 minutes. What is the temperature of
the surroundings?​

Answer 1

Answer:

390°

Explanation:

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Answer 2

Answer:

1st case, change in temperature,

$dT= 70 - 55 = 15°C$

time interval, dt=5 min <br> Average temperature,

$T = \frac{70 + 55}{2} = 62.5°C$

Temperature of surrounding,

$T _{0} = ?$

As,

$\frac{dT}{dt} = - K(T - T _{0})$

$\frac{15}{5} = -K (62.5 - T _{0})$

.....(i) 2nd case,

$dT = 55 - 45 = 10°C$

$dt = 10 - 5 = 5mins$

$T= \frac{55 + 45}{2} = 50°C$

Then,

$\frac{10}{5} = - K(50 - T_{0})$

....(ii)

Dividing (i) by (ii) and solving we get,

$T _{0} = 25°C$

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