Thermal Properties of Matter NEET PYQ – Previous Year Questions with Solutions for NEET Physics

Q 1 :
A cup of coffee cools from $90 ° C$ to $80 ° C$ in $t$ minutes, when the room temperature is $20 ° C$ . The time taken by a similar cup of coffee to cool from $80 ° C$ to $60 ° C$ at a room temperature same at $20 ° C$ is [2021]

$\frac{5}{13} t$
$\frac{13}{10} t$
$\frac{13}{5} t$
$\frac{10}{13} t$

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(3)

Initial temperature $(T_{i}) = 90 ° C$

Final temperature $(T_{f}) = 80 ° C$

Room temperature $(T_{0}) = 20 ° C$

Let time taken be $t$ minutes.

According to Newton’s law of cooling,

Rate of cooling $\frac{d T}{d t} = K [\frac{(T_{i} + T_{f})}{2} - T_{0}]$

$\frac{(T_{f} - T_{i})}{t} = K [\frac{(90 + 80)}{2} - 20] \Rightarrow \frac{90 - 80}{t} = K [65]$

$K = \frac{10}{65 t}$

In 2nd condition,

Initial temperature $T_{i} = 80 ° C$

Final temperature $T_{f} = 60 ° C$

Let time taken be $t'$ minutes.

Then, $\frac{(80 - 60)}{t'} = \frac{10}{65 t} [\frac{(80 + 60)}{2} - 20]$ $\frac{20}{t'} = \frac{10}{65 t} (50)$

$\Rightarrow t' = \frac{13}{5} t$

Q 2 :
A body cools from a temperature $3 T$ to $2 T$ in 10 minutes. The room temperature is $T$ . Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be [2016]

$\frac{7}{4} T$
$\frac{3}{2} T$
$\frac{4}{3} T$
$T$

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(2)

According to Newton’s law of cooling,

$\frac{d T}{d t} = K (T - T_{s})$

For two cases, $\frac{d T_{1}}{d t} = K (T_{1} - T_{s}) and \frac{d T_{2}}{d t} = K (T_{2} - T_{s})$

Here, $T_{s} = T,$ $T_{1} = \frac{3 T + 2 T}{2} = 2.5 T and \frac{d T_{1}}{d t} = \frac{3 T - 2 T}{10} = \frac{T}{10}$

$T_{2} = \frac{2 T + T'}{2} and \frac{d T_{2}}{d t} = \frac{2 T - T'}{10}$

So, $\frac{T}{10} = K (2.5 T - T)$ ...(i)

and $\frac{2 T - T'}{10} = K (\frac{2 T + T'}{2} - T)$ ...(ii)

Dividing eqn. (i) by eqn. (ii), we get

$\frac{T}{2 T - T'} = \frac{(2.5 T - T)}{(\frac{2 T + T'}{2} - T)} or \frac{2 T + T'}{2} - T = (2 T - T') \times \frac{3}{2}$

$T' = 3 (2 T - T') or, 4 T' = 6 T ∴ T' = \frac{3}{2} T$

Q 3 :
Certain quantity of water cools from $70 ° C$ to $60 ° C$ in the first 5 minutes and to $54 ° C$ in the next 5 minutes. The temperature of the surroundings is [2014]

$45 ° C$
$20 ° C$
$42 ° C$
$10 ° C$

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(1)

Let $T_{s}$ be the temperature of the surroundings.

According to Newton’s law of cooling

$\frac{T_{1} - T_{2}}{t} = K (\frac{T_{1} + T_{2}}{2} - T_{s})$

For first 5 minutes,

$T_{1} = 70 ° C,$ $T_{2} = 60 ° C,$ $t = 5$ minutes

$∴$ $\frac{70 - 60}{5} = K (\frac{70 + 60}{2} - T_{s}) = K (65 - T_{s})$ ...(i)

For next 5 minutes,

$T_{1} = 60 ° C,$ $T_{2} = 54 ° C,$ $t = 5$ minutes

$∴$ $\frac{60 - 54}{5} = K (\frac{60 + 54}{2} - T_{s})$

$\frac{6}{5} = K (57 - T_{s})$ ...(ii)

Divide eqn. (i) by eqn. (ii), we get

$\frac{5}{3} = \frac{65 - T_{s}}{57 - T_{s}}$

$285 - 5 T_{s} = 195 - 3 T_{s}$

$2 T_{s} = 90 \Rightarrow T_{s} = 45 ° C$

Topic Question Set

Q 1 :
A cup of coffee cools from $90 ° C$ to $80 ° C$ in $t$ minutes, when the room temperature is $20 ° C$ . The time taken by a similar cup of coffee to cool from $80 ° C$ to $60 ° C$ at a room temperature same at $20 ° C$ is [2021]

Q 2 :
A body cools from a temperature $3 T$ to $2 T$ in 10 minutes. The room temperature is $T$ . Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be [2016]

Q 3 :
Certain quantity of water cools from $70 ° C$ to $60 ° C$ in the first 5 minutes and to $54 ° C$ in the next 5 minutes. The temperature of the surroundings is [2014]

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Topic Question Set

Q 1 : A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is [2021]

Q 2 : A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be [2016]

Q 3 : Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next 5 minutes. The temperature of the surroundings is [2014]

Want Us to Email you About Special Offers & Updates?

Q 1 :
A cup of coffee cools from $90 ° C$ to $80 ° C$ in $t$ minutes, when the room temperature is $20 ° C$ . The time taken by a similar cup of coffee to cool from $80 ° C$ to $60 ° C$ at a room temperature same at $20 ° C$ is [2021]

Q 2 :
A body cools from a temperature $3 T$ to $2 T$ in 10 minutes. The room temperature is $T$ . Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be [2016]

Q 3 :
Certain quantity of water cools from $70 ° C$ to $60 ° C$ in the first 5 minutes and to $54 ° C$ in the next 5 minutes. The temperature of the surroundings is [2014]