QBManager

PHXI11:THERMAL PROPERTIES OF MATTER

366644 The temperature of a body falls from $52^{\circ} C$ to $36^{\circ} C$ in 10 minutes when placed in a surrounding of constant temperature $20^{\circ} C$ . What will be the temperature of the body after another 10 min? (Use Newton's law of cooling)

1

28^{\circ} C

2

20^{\circ} C

3

32^{\circ} C

4

24^{\circ} C

Explanation:

Applying newton's law of cooling
$\begin{aligned} \frac{52 - 36}{10 min} = \frac{- b}{m c} [\frac{52 + 36}{2} - 20] \\ \Rightarrow \frac{36 - θ}{10 min} = \frac{- b}{m c} [\frac{36 + θ}{2} - 20] \\ \Rightarrow θ = 28^{\circ} C \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366645 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the room temperature is $20^{\circ} C$ . How long will it cool from $74^{\circ} C$ to $66^{\circ} C$ ?

1 2 minutes

2 2.8 minutes

3 2.5 minutes

4 1.8 minutes

Explanation:

According to Newton's law of cooling
$\frac{Δ θ}{Δ t} = k (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
case-1
$θ_{1} = 94^{\circ} C, θ_{2} = 86^{\circ} C, θ_{0} = 20^{\circ} C$
$\frac{8}{2} = k (\frac{94 + 86}{2} - 20) = 70 k$
$k = \frac{4}{70}$
Case - (2)
$θ_{1} = 74^{\circ} C, θ_{2} = 66^{\circ} C$
$\frac{74 - 66}{t} = \frac{4}{70} (\frac{74 + 66}{2} - 20) = \frac{4}{70} \times 50$
$t = \frac{8 \times 7}{4 \times 5} = 2.8 m i n u t e s$

KCET - 2018

PHXI11:THERMAL PROPERTIES OF MATTER

366646 The rates of cooling of two different liquids put in exactly similar surroundings are the same if

1 Equal masses of the liquids at the same temperature are taken

2 The masses of the liquid are equal

3 Heat capacities of the liquids must be same

4 None of the above

Explanation:

$\frac{d T}{d t} α \frac{σ A}{m c} (T^{4} - T_{0}^{4})$ . If the liquids put in exactly similar surrounding then the rate of cooling will be same if $A$ , $m$ , $c$ & $T$ must be same or $\frac{A}{m c} (T^{4} - T_{0}^{4})$ must be same for both

PHXI11:THERMAL PROPERTIES OF MATTER

366647 A body cools from $80^{\circ} C$ to $60^{\circ} C$ in 5 minutes.The temperature of the surrounding is $20^{\circ} C$ .The time it takes to cool from $60^{\circ} C$ to $40^{\circ} C$ is

1

420 s

2

500 s

3

450 s

4

\frac{25}{3} s

Explanation:

Initial temperature $(T_{i}) = 80^{\circ} C$
Final temperature $(T_{f}) = 60^{\circ} C$
Temperature of the surrounding $(T_{0}) = 20^{\circ} C$
Time, $t = 5 m i n$
According to Newton's law of cooling
$- \frac{d T}{d t} = K (T - T_{0})$ . Here $T = \frac{1}{2} (T_{i} + T_{f})$ .
Rate of cooling, $= K [\frac{(T_{i} + T_{f})}{2} - T_{0}]$
$\frac{(T_{f} - T_{i})}{t} = K [\frac{80 + 60}{2} - 20]$
$\frac{80 - 60}{5} = K [70 - 20]$ $4 = K [50]; K = \frac{4}{50} = \frac{2}{25}$
In second condition,
Initial temperature $= T_{i} = 60^{\circ} C$
Final temperature $= T_{f} = 40^{\circ} C$
Time taken for cooling is $t$ .
According to Newton's law of cooling,
$\frac{60 - 40}{t} = \frac{2}{25} [\frac{60 + 40}{2} - 20];$
$\frac{20}{t} = \frac{2}{25} [30]$
$\Rightarrow t = 500 s$ .

JEE - 2023

PHXI11:THERMAL PROPERTIES OF MATTER

366644 The temperature of a body falls from $52^{\circ} C$ to $36^{\circ} C$ in 10 minutes when placed in a surrounding of constant temperature $20^{\circ} C$ . What will be the temperature of the body after another 10 min? (Use Newton's law of cooling)

1

28^{\circ} C

2

20^{\circ} C

3

32^{\circ} C

4

24^{\circ} C

Explanation:

Applying newton's law of cooling
$\begin{aligned} \frac{52 - 36}{10 min} = \frac{- b}{m c} [\frac{52 + 36}{2} - 20] \\ \Rightarrow \frac{36 - θ}{10 min} = \frac{- b}{m c} [\frac{36 + θ}{2} - 20] \\ \Rightarrow θ = 28^{\circ} C \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366645 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the room temperature is $20^{\circ} C$ . How long will it cool from $74^{\circ} C$ to $66^{\circ} C$ ?

1 2 minutes

2 2.8 minutes

3 2.5 minutes

4 1.8 minutes

Explanation:

According to Newton's law of cooling
$\frac{Δ θ}{Δ t} = k (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
case-1
$θ_{1} = 94^{\circ} C, θ_{2} = 86^{\circ} C, θ_{0} = 20^{\circ} C$
$\frac{8}{2} = k (\frac{94 + 86}{2} - 20) = 70 k$
$k = \frac{4}{70}$
Case - (2)
$θ_{1} = 74^{\circ} C, θ_{2} = 66^{\circ} C$
$\frac{74 - 66}{t} = \frac{4}{70} (\frac{74 + 66}{2} - 20) = \frac{4}{70} \times 50$
$t = \frac{8 \times 7}{4 \times 5} = 2.8 m i n u t e s$

KCET - 2018

PHXI11:THERMAL PROPERTIES OF MATTER

366646 The rates of cooling of two different liquids put in exactly similar surroundings are the same if

1 Equal masses of the liquids at the same temperature are taken

2 The masses of the liquid are equal

3 Heat capacities of the liquids must be same

4 None of the above

Explanation:

$\frac{d T}{d t} α \frac{σ A}{m c} (T^{4} - T_{0}^{4})$ . If the liquids put in exactly similar surrounding then the rate of cooling will be same if $A$ , $m$ , $c$ & $T$ must be same or $\frac{A}{m c} (T^{4} - T_{0}^{4})$ must be same for both

PHXI11:THERMAL PROPERTIES OF MATTER

366647 A body cools from $80^{\circ} C$ to $60^{\circ} C$ in 5 minutes.The temperature of the surrounding is $20^{\circ} C$ .The time it takes to cool from $60^{\circ} C$ to $40^{\circ} C$ is

1

420 s

2

500 s

3

450 s

4

\frac{25}{3} s

Explanation:

Initial temperature $(T_{i}) = 80^{\circ} C$
Final temperature $(T_{f}) = 60^{\circ} C$
Temperature of the surrounding $(T_{0}) = 20^{\circ} C$
Time, $t = 5 m i n$
According to Newton's law of cooling
$- \frac{d T}{d t} = K (T - T_{0})$ . Here $T = \frac{1}{2} (T_{i} + T_{f})$ .
Rate of cooling, $= K [\frac{(T_{i} + T_{f})}{2} - T_{0}]$
$\frac{(T_{f} - T_{i})}{t} = K [\frac{80 + 60}{2} - 20]$
$\frac{80 - 60}{5} = K [70 - 20]$ $4 = K [50]; K = \frac{4}{50} = \frac{2}{25}$
In second condition,
Initial temperature $= T_{i} = 60^{\circ} C$
Final temperature $= T_{f} = 40^{\circ} C$
Time taken for cooling is $t$ .
According to Newton's law of cooling,
$\frac{60 - 40}{t} = \frac{2}{25} [\frac{60 + 40}{2} - 20];$
$\frac{20}{t} = \frac{2}{25} [30]$
$\Rightarrow t = 500 s$ .

JEE - 2023

PHXI11:THERMAL PROPERTIES OF MATTER

366644 The temperature of a body falls from $52^{\circ} C$ to $36^{\circ} C$ in 10 minutes when placed in a surrounding of constant temperature $20^{\circ} C$ . What will be the temperature of the body after another 10 min? (Use Newton's law of cooling)

1

28^{\circ} C

2

20^{\circ} C

3

32^{\circ} C

4

24^{\circ} C

Explanation:

Applying newton's law of cooling
$\begin{aligned} \frac{52 - 36}{10 min} = \frac{- b}{m c} [\frac{52 + 36}{2} - 20] \\ \Rightarrow \frac{36 - θ}{10 min} = \frac{- b}{m c} [\frac{36 + θ}{2} - 20] \\ \Rightarrow θ = 28^{\circ} C \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366645 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the room temperature is $20^{\circ} C$ . How long will it cool from $74^{\circ} C$ to $66^{\circ} C$ ?

1 2 minutes

2 2.8 minutes

3 2.5 minutes

4 1.8 minutes

Explanation:

According to Newton's law of cooling
$\frac{Δ θ}{Δ t} = k (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
case-1
$θ_{1} = 94^{\circ} C, θ_{2} = 86^{\circ} C, θ_{0} = 20^{\circ} C$
$\frac{8}{2} = k (\frac{94 + 86}{2} - 20) = 70 k$
$k = \frac{4}{70}$
Case - (2)
$θ_{1} = 74^{\circ} C, θ_{2} = 66^{\circ} C$
$\frac{74 - 66}{t} = \frac{4}{70} (\frac{74 + 66}{2} - 20) = \frac{4}{70} \times 50$
$t = \frac{8 \times 7}{4 \times 5} = 2.8 m i n u t e s$

KCET - 2018

PHXI11:THERMAL PROPERTIES OF MATTER

366646 The rates of cooling of two different liquids put in exactly similar surroundings are the same if

1 Equal masses of the liquids at the same temperature are taken

2 The masses of the liquid are equal

3 Heat capacities of the liquids must be same

4 None of the above

Explanation:

$\frac{d T}{d t} α \frac{σ A}{m c} (T^{4} - T_{0}^{4})$ . If the liquids put in exactly similar surrounding then the rate of cooling will be same if $A$ , $m$ , $c$ & $T$ must be same or $\frac{A}{m c} (T^{4} - T_{0}^{4})$ must be same for both

PHXI11:THERMAL PROPERTIES OF MATTER

366647 A body cools from $80^{\circ} C$ to $60^{\circ} C$ in 5 minutes.The temperature of the surrounding is $20^{\circ} C$ .The time it takes to cool from $60^{\circ} C$ to $40^{\circ} C$ is

1

420 s

2

500 s

3

450 s

4

\frac{25}{3} s

Explanation:

Initial temperature $(T_{i}) = 80^{\circ} C$
Final temperature $(T_{f}) = 60^{\circ} C$
Temperature of the surrounding $(T_{0}) = 20^{\circ} C$
Time, $t = 5 m i n$
According to Newton's law of cooling
$- \frac{d T}{d t} = K (T - T_{0})$ . Here $T = \frac{1}{2} (T_{i} + T_{f})$ .
Rate of cooling, $= K [\frac{(T_{i} + T_{f})}{2} - T_{0}]$
$\frac{(T_{f} - T_{i})}{t} = K [\frac{80 + 60}{2} - 20]$
$\frac{80 - 60}{5} = K [70 - 20]$ $4 = K [50]; K = \frac{4}{50} = \frac{2}{25}$
In second condition,
Initial temperature $= T_{i} = 60^{\circ} C$
Final temperature $= T_{f} = 40^{\circ} C$
Time taken for cooling is $t$ .
According to Newton's law of cooling,
$\frac{60 - 40}{t} = \frac{2}{25} [\frac{60 + 40}{2} - 20];$
$\frac{20}{t} = \frac{2}{25} [30]$
$\Rightarrow t = 500 s$ .

JEE - 2023

PHXI11:THERMAL PROPERTIES OF MATTER

366644 The temperature of a body falls from $52^{\circ} C$ to $36^{\circ} C$ in 10 minutes when placed in a surrounding of constant temperature $20^{\circ} C$ . What will be the temperature of the body after another 10 min? (Use Newton's law of cooling)

1

28^{\circ} C

2

20^{\circ} C

3

32^{\circ} C

4

24^{\circ} C

Explanation:

Applying newton's law of cooling
$\begin{aligned} \frac{52 - 36}{10 min} = \frac{- b}{m c} [\frac{52 + 36}{2} - 20] \\ \Rightarrow \frac{36 - θ}{10 min} = \frac{- b}{m c} [\frac{36 + θ}{2} - 20] \\ \Rightarrow θ = 28^{\circ} C \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366645 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the room temperature is $20^{\circ} C$ . How long will it cool from $74^{\circ} C$ to $66^{\circ} C$ ?

1 2 minutes

2 2.8 minutes

3 2.5 minutes

4 1.8 minutes

Explanation:

According to Newton's law of cooling
$\frac{Δ θ}{Δ t} = k (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
case-1
$θ_{1} = 94^{\circ} C, θ_{2} = 86^{\circ} C, θ_{0} = 20^{\circ} C$
$\frac{8}{2} = k (\frac{94 + 86}{2} - 20) = 70 k$
$k = \frac{4}{70}$
Case - (2)
$θ_{1} = 74^{\circ} C, θ_{2} = 66^{\circ} C$
$\frac{74 - 66}{t} = \frac{4}{70} (\frac{74 + 66}{2} - 20) = \frac{4}{70} \times 50$
$t = \frac{8 \times 7}{4 \times 5} = 2.8 m i n u t e s$

KCET - 2018

PHXI11:THERMAL PROPERTIES OF MATTER

366646 The rates of cooling of two different liquids put in exactly similar surroundings are the same if

1 Equal masses of the liquids at the same temperature are taken

2 The masses of the liquid are equal

3 Heat capacities of the liquids must be same

4 None of the above

Explanation:

$\frac{d T}{d t} α \frac{σ A}{m c} (T^{4} - T_{0}^{4})$ . If the liquids put in exactly similar surrounding then the rate of cooling will be same if $A$ , $m$ , $c$ & $T$ must be same or $\frac{A}{m c} (T^{4} - T_{0}^{4})$ must be same for both

PHXI11:THERMAL PROPERTIES OF MATTER

366647 A body cools from $80^{\circ} C$ to $60^{\circ} C$ in 5 minutes.The temperature of the surrounding is $20^{\circ} C$ .The time it takes to cool from $60^{\circ} C$ to $40^{\circ} C$ is

1

420 s

2

500 s

3

450 s

4

\frac{25}{3} s

Explanation:

Initial temperature $(T_{i}) = 80^{\circ} C$
Final temperature $(T_{f}) = 60^{\circ} C$
Temperature of the surrounding $(T_{0}) = 20^{\circ} C$
Time, $t = 5 m i n$
According to Newton's law of cooling
$- \frac{d T}{d t} = K (T - T_{0})$ . Here $T = \frac{1}{2} (T_{i} + T_{f})$ .
Rate of cooling, $= K [\frac{(T_{i} + T_{f})}{2} - T_{0}]$
$\frac{(T_{f} - T_{i})}{t} = K [\frac{80 + 60}{2} - 20]$
$\frac{80 - 60}{5} = K [70 - 20]$ $4 = K [50]; K = \frac{4}{50} = \frac{2}{25}$
In second condition,
Initial temperature $= T_{i} = 60^{\circ} C$
Final temperature $= T_{f} = 40^{\circ} C$
Time taken for cooling is $t$ .
According to Newton's law of cooling,
$\frac{60 - 40}{t} = \frac{2}{25} [\frac{60 + 40}{2} - 20];$
$\frac{20}{t} = \frac{2}{25} [30]$
$\Rightarrow t = 500 s$ .

JEE - 2023