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+1 vote
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in Physics by (30.7k points)
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A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The ti me taken by a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C.

(1) 13/10 t

(2) 13/5 t

(3) 10/13 t

(4) 5/13 t

2 Answers

+1 vote
by (15.0k points)
selected by
 
Best answer

Correct option is (2) 13/5 t

By using Newton's law of cooling, 

\(\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}2 - T_s\right)\)     .......(1)

For the first case we have;

Given: Initial Temperature, T1 = 90°C

Final temperature, T= 80°C

and surrounding temperature, Ts = 20°C

Now, on putting the above values in equation (1) we get;

\(\frac{90 - 80}t= k \left(\frac{90 + 80}2 - 20\right)\)

⇒ \(k[85 - 20] = \frac{90 - 80}t\)

⇒ \(k[65] = \frac{10}t\)

⇒ \(k = \frac 2{13 t}\)    ......(2)

For the second case we have;

Similarly, 

Given: Initial Temperature, T1 = 80°C

Final temperature, T2 = 60°C

and surrounding temperature, Ts = 20°C

\(\frac{80 - 60}{t'} = k\left(\frac{80 + 60}2 - 20\right)\)

⇒ \(k[50] = \frac{20}{t'}\)   ......(3)

where t' is the time taken in the second case

Substituting the equation (2) in (3) we get;

\(\frac 2{13t} (50) = \frac{20}{t'} \)

⇒ \(t' = \frac{13 t}5\)

+4 votes
by (30.6k points)

Option : (2) 13/5 t

According to Newton's law of cooling,

\(\frac{T_1-T_2}{t}\) = K\([\frac{T_1+T_2}{2}-T_0]\)

For 1st cup of coffee,

⇒ \(\frac{90-80}{t}\) = K\([\frac{90+80}{2}-20]\) ..(1)

For 2nd cup of coffee,

⇒ \(\frac{80-60}{t'}\) = K\([\frac{80+60}{2}-20]\) ..(2)

Divide (1) by (2),

\(\frac{t'}{2t}\) \(\frac{65}{50}\)

⇒ t' = \(\frac{13}{5}t\)

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