Question

A body cools from 100 to 90 in 10 minutes and from 90 to 80 in 15 minutes.the temperature of surrounding is

Answer 1

may be the temperature if the surrounding is equal to the body

Answer 2

Answer:

The temperature of the surroundings is 65°C.

Explanation:

We can deduce the following from Newton's rule of cooling:

$\frac{T_2-T_1}{t}=k{(\frac{T_1+T_2}{2}-T_0)$      (1)

Where,

T₁, T₂=Respective temperatures

T₀=temperature of the surrounding

t=time during which the change occurs

k=constant

Case I:

T₁=100°C

T₂=90°C

t=10 minutes

By inserting them in equation (1) we get;

$\frac{100-90}{10}=k{(\frac{100+90}{2}-T_0)$

$1=k(95-T_0)$     (2)

Case II:

T₁=90°C

T₂=80°C

t=15 minutes

By inserting them in equation (1) we get;

$\frac{90-80}{15}=k{(\frac{90+80}{2}-T_0)$

$\frac{10}{15}=k{(85-T_0)$

$\frac{2}{3}=k{(85-T_0)$

$2=3k{(85-T_0)$        (3)

We can get the following result by taking the ratio of equations (2) and (3).

$\frac{1}{2}=\frac{k(95-T_0)}{3k(85-T_0)}$

$\frac{1}{2}=\frac{(95-T_0)}{3(85-T_0)}$

$3(85-T_0)=2(95-T_0)$

$255-3T_0=190-2T_0$

$3T_0-2T_0=255-190$

$T_0=65\textdegree C$

Hence, the temperature of the surroundings is 65°C.

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