QBManager

Heat Transfer

149647 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2 \min$ , when the room temperature is $20^{\circ} C$ . The time taken for the food to cool from $86^{\circ} C$ to $74^{\circ} C$ will be

1

500 s

2

420 s

3

200 s

4

210 s

Explanation:

D Given, $T_{1} = 94^{\circ} C, T_{2} = 86^{\circ} C, t = 2 \min$
According to Newton's law of cooling-
$\frac{T_{1} - T_{2}}{2} = K [\frac{T_{1} + T_{2}}{2} - T_{s}]$
Where, $T_{s}$ is the surrounding temperature which is $20^{\circ} C$ . For first case,
$\frac{94 - 86}{2 \times 60} = K [\frac{94 + 86}{2} - 20]$
$\frac{8}{2 \times 60} = K \times 70$
$\frac{1}{15} = K \times 70$
For second case,
$\frac{86 - 74}{t} = K [\frac{86 + 74}{2} - 20]$
$\frac{12}{t} = K \times 60$
Dividing equation (i) by equation (ii), we get
$\frac{\frac{1}{15}}{\frac{12}{t}} = \frac{K \times 70}{K \times 60}$
$\frac{t}{12 \times 15} = \frac{7}{6}$
$t = 30 \times 7$
$t = 210 s$

AIIMS-27.05.2018(M)]

Heat Transfer

149648 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2 \min$ . When the room temperature is $20^{\circ} C$ . How long will it cool from $74^{0} C$ to $66^{\circ} C$ ?

1

2 \min

2

2.8 \min

3

2.5 \min

4

1.8 \min

Explanation:

B According to Newton's law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K [\frac{θ_{1} + θ_{2}}{2} - θ_{o}]$
For case - (I)
$t = 2 \min$
$θ_{1} - θ_{2} = 94 - 86 = 8^{\circ} C$
$θ_{o} = 20^{\circ} C$
$\frac{θ_{1} + θ_{2}}{2} = \frac{94 + 86}{2} = \frac{180}{2} = 90^{\circ} C$
So, put these value in Newton's law of cooling formula we get
$\frac{8}{2} = K (90 - 20)$
$K = \frac{4}{70}$
For case - (II)
$θ_{1} - θ_{2} = 74 - 66 = 8^{\circ} C$
$t =$ ?
$\frac{θ_{1} + θ_{2}}{2} = \frac{74 + 66}{2} = \frac{140}{2} = 70^{\circ} C$
$θ_{o} = 20^{\circ} C$
Thus, using Newton's cooling law,
$\frac{8}{t} = K (70^{\circ} - 20^{\circ})$ Put the value of $K$ we get,
$\frac{8}{t} = \frac{4}{70} (50^{\circ})$
$t = \frac{70}{25} = 2.8$
Thus, it will take $2.8 \min$ .

Karnataka CET-2016

Heat Transfer

149649 Hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes when the room temperature is $20^{\circ} C$ . How long would the food take to cool from $71^{\circ} C$ to $69^{\circ} C$ ?

1

12 \sec

2

25 \sec

3

16 \sec

4

42 \sec

Explanation:

D We know Newton's law of cooling -
$\frac{T_{1} - T_{2}}{t} = - K (\frac{T_{1} + T_{2}}{2} - T_{S})$
For case-I
$T_{1} = 94^{\circ} C, T_{2} = 86^{\circ} C, t = 2 \min$ and $T_{s} = 20^{\circ} C$
$\frac{94 - 86}{2} = - K (\frac{94 + 86}{2} - 20)$
$K = - (4 / 70) = - (2 / 35)$
For case - II
$T_{1} = 71^{\circ} C, T_{2} = 69^{\circ} C, T_{s} = 20^{\circ} C$ and $t =$ ?
From equation (i)
$\frac{71 - 69}{t} = - (- \frac{2}{35}) \times (\frac{71 + 69}{2} - 20)$
$t = 0.7 min$
$t = 0.7 \times 60$
$t = 42 second$

AMU-2012

Heat Transfer

149650 A body cools from $60^{\circ} C$ to $50^{\circ} C$ in $10 \min$ . If room temperature is $25^{\circ} C$ of body at the end of next 10 min, will be

1

{38.5}^{\circ} C

2

40^{\circ} C

3

45^{\circ} C

4

{42.85}^{\circ} C

Explanation:

D Given,
Initial Temp $(θ_{1}) = 60^{\circ} C$ , Final Temp. $(θ_{2}) = 50^{\circ} C$
Atmospheric Temp. $(θ) = 25^{\circ} C$
Time $(t) = 10$ min
According to Newton's law,
$\frac{θ_{1} - θ_{2}}{t} = - K (\frac{θ_{1} + θ_{2}}{2} - θ)$
$\frac{60 - 50}{10} = - K (\frac{60 + 50}{2} - 25)$
$1 = - K (55 - 25)$
$K = - \frac{1}{30}$
Case-II
Again using Newton's law cooling,
$\frac{50 - θ_{3}}{10} = - (- \frac{1}{30}) (\frac{50 + θ_{3}}{2} - 25)$
$150 - 3 θ_{3} = \frac{50 + θ_{3} - 50}{2}$
$300 - 6 θ_{3} = θ_{3}$
$7 θ_{3} = 300$
$θ_{3} = {42.85}^{\circ} C$

UP CPMT-2012

Heat Transfer

149647 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2 \min$ , when the room temperature is $20^{\circ} C$ . The time taken for the food to cool from $86^{\circ} C$ to $74^{\circ} C$ will be

1

500 s

2

420 s

3

200 s

4

210 s

Explanation:

D Given, $T_{1} = 94^{\circ} C, T_{2} = 86^{\circ} C, t = 2 \min$
According to Newton's law of cooling-
$\frac{T_{1} - T_{2}}{2} = K [\frac{T_{1} + T_{2}}{2} - T_{s}]$
Where, $T_{s}$ is the surrounding temperature which is $20^{\circ} C$ . For first case,
$\frac{94 - 86}{2 \times 60} = K [\frac{94 + 86}{2} - 20]$
$\frac{8}{2 \times 60} = K \times 70$
$\frac{1}{15} = K \times 70$
For second case,
$\frac{86 - 74}{t} = K [\frac{86 + 74}{2} - 20]$
$\frac{12}{t} = K \times 60$
Dividing equation (i) by equation (ii), we get
$\frac{\frac{1}{15}}{\frac{12}{t}} = \frac{K \times 70}{K \times 60}$
$\frac{t}{12 \times 15} = \frac{7}{6}$
$t = 30 \times 7$
$t = 210 s$

AIIMS-27.05.2018(M)]

Heat Transfer

149648 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2 \min$ . When the room temperature is $20^{\circ} C$ . How long will it cool from $74^{0} C$ to $66^{\circ} C$ ?

1

2 \min

2

2.8 \min

3

2.5 \min

4

1.8 \min

Explanation:

B According to Newton's law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K [\frac{θ_{1} + θ_{2}}{2} - θ_{o}]$
For case - (I)
$t = 2 \min$
$θ_{1} - θ_{2} = 94 - 86 = 8^{\circ} C$
$θ_{o} = 20^{\circ} C$
$\frac{θ_{1} + θ_{2}}{2} = \frac{94 + 86}{2} = \frac{180}{2} = 90^{\circ} C$
So, put these value in Newton's law of cooling formula we get
$\frac{8}{2} = K (90 - 20)$
$K = \frac{4}{70}$
For case - (II)
$θ_{1} - θ_{2} = 74 - 66 = 8^{\circ} C$
$t =$ ?
$\frac{θ_{1} + θ_{2}}{2} = \frac{74 + 66}{2} = \frac{140}{2} = 70^{\circ} C$
$θ_{o} = 20^{\circ} C$
Thus, using Newton's cooling law,
$\frac{8}{t} = K (70^{\circ} - 20^{\circ})$ Put the value of $K$ we get,
$\frac{8}{t} = \frac{4}{70} (50^{\circ})$
$t = \frac{70}{25} = 2.8$
Thus, it will take $2.8 \min$ .

Karnataka CET-2016

Heat Transfer

149649 Hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes when the room temperature is $20^{\circ} C$ . How long would the food take to cool from $71^{\circ} C$ to $69^{\circ} C$ ?

1

12 \sec

2

25 \sec

3

16 \sec

4

42 \sec

Explanation:

D We know Newton's law of cooling -
$\frac{T_{1} - T_{2}}{t} = - K (\frac{T_{1} + T_{2}}{2} - T_{S})$
For case-I
$T_{1} = 94^{\circ} C, T_{2} = 86^{\circ} C, t = 2 \min$ and $T_{s} = 20^{\circ} C$
$\frac{94 - 86}{2} = - K (\frac{94 + 86}{2} - 20)$
$K = - (4 / 70) = - (2 / 35)$
For case - II
$T_{1} = 71^{\circ} C, T_{2} = 69^{\circ} C, T_{s} = 20^{\circ} C$ and $t =$ ?
From equation (i)
$\frac{71 - 69}{t} = - (- \frac{2}{35}) \times (\frac{71 + 69}{2} - 20)$
$t = 0.7 min$
$t = 0.7 \times 60$
$t = 42 second$

AMU-2012

Heat Transfer

149650 A body cools from $60^{\circ} C$ to $50^{\circ} C$ in $10 \min$ . If room temperature is $25^{\circ} C$ of body at the end of next 10 min, will be

1

{38.5}^{\circ} C

2

40^{\circ} C

3

45^{\circ} C

4

{42.85}^{\circ} C

Explanation:

D Given,
Initial Temp $(θ_{1}) = 60^{\circ} C$ , Final Temp. $(θ_{2}) = 50^{\circ} C$
Atmospheric Temp. $(θ) = 25^{\circ} C$
Time $(t) = 10$ min
According to Newton's law,
$\frac{θ_{1} - θ_{2}}{t} = - K (\frac{θ_{1} + θ_{2}}{2} - θ)$
$\frac{60 - 50}{10} = - K (\frac{60 + 50}{2} - 25)$
$1 = - K (55 - 25)$
$K = - \frac{1}{30}$
Case-II
Again using Newton's law cooling,
$\frac{50 - θ_{3}}{10} = - (- \frac{1}{30}) (\frac{50 + θ_{3}}{2} - 25)$
$150 - 3 θ_{3} = \frac{50 + θ_{3} - 50}{2}$
$300 - 6 θ_{3} = θ_{3}$
$7 θ_{3} = 300$
$θ_{3} = {42.85}^{\circ} C$

UP CPMT-2012

Heat Transfer

149647 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2 \min$ , when the room temperature is $20^{\circ} C$ . The time taken for the food to cool from $86^{\circ} C$ to $74^{\circ} C$ will be

1

500 s

2

420 s

3

200 s

4

210 s

Explanation:

D Given, $T_{1} = 94^{\circ} C, T_{2} = 86^{\circ} C, t = 2 \min$
According to Newton's law of cooling-
$\frac{T_{1} - T_{2}}{2} = K [\frac{T_{1} + T_{2}}{2} - T_{s}]$
Where, $T_{s}$ is the surrounding temperature which is $20^{\circ} C$ . For first case,
$\frac{94 - 86}{2 \times 60} = K [\frac{94 + 86}{2} - 20]$
$\frac{8}{2 \times 60} = K \times 70$
$\frac{1}{15} = K \times 70$
For second case,
$\frac{86 - 74}{t} = K [\frac{86 + 74}{2} - 20]$
$\frac{12}{t} = K \times 60$
Dividing equation (i) by equation (ii), we get
$\frac{\frac{1}{15}}{\frac{12}{t}} = \frac{K \times 70}{K \times 60}$
$\frac{t}{12 \times 15} = \frac{7}{6}$
$t = 30 \times 7$
$t = 210 s$

AIIMS-27.05.2018(M)]

Heat Transfer

149648 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2 \min$ . When the room temperature is $20^{\circ} C$ . How long will it cool from $74^{0} C$ to $66^{\circ} C$ ?

1

2 \min

2

2.8 \min

3

2.5 \min

4

1.8 \min

Explanation:

B According to Newton's law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K [\frac{θ_{1} + θ_{2}}{2} - θ_{o}]$
For case - (I)
$t = 2 \min$
$θ_{1} - θ_{2} = 94 - 86 = 8^{\circ} C$
$θ_{o} = 20^{\circ} C$
$\frac{θ_{1} + θ_{2}}{2} = \frac{94 + 86}{2} = \frac{180}{2} = 90^{\circ} C$
So, put these value in Newton's law of cooling formula we get
$\frac{8}{2} = K (90 - 20)$
$K = \frac{4}{70}$
For case - (II)
$θ_{1} - θ_{2} = 74 - 66 = 8^{\circ} C$
$t =$ ?
$\frac{θ_{1} + θ_{2}}{2} = \frac{74 + 66}{2} = \frac{140}{2} = 70^{\circ} C$
$θ_{o} = 20^{\circ} C$
Thus, using Newton's cooling law,
$\frac{8}{t} = K (70^{\circ} - 20^{\circ})$ Put the value of $K$ we get,
$\frac{8}{t} = \frac{4}{70} (50^{\circ})$
$t = \frac{70}{25} = 2.8$
Thus, it will take $2.8 \min$ .

Karnataka CET-2016

Heat Transfer

149649 Hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes when the room temperature is $20^{\circ} C$ . How long would the food take to cool from $71^{\circ} C$ to $69^{\circ} C$ ?

1

12 \sec

2

25 \sec

3

16 \sec

4

42 \sec

Explanation:

D We know Newton's law of cooling -
$\frac{T_{1} - T_{2}}{t} = - K (\frac{T_{1} + T_{2}}{2} - T_{S})$
For case-I
$T_{1} = 94^{\circ} C, T_{2} = 86^{\circ} C, t = 2 \min$ and $T_{s} = 20^{\circ} C$
$\frac{94 - 86}{2} = - K (\frac{94 + 86}{2} - 20)$
$K = - (4 / 70) = - (2 / 35)$
For case - II
$T_{1} = 71^{\circ} C, T_{2} = 69^{\circ} C, T_{s} = 20^{\circ} C$ and $t =$ ?
From equation (i)
$\frac{71 - 69}{t} = - (- \frac{2}{35}) \times (\frac{71 + 69}{2} - 20)$
$t = 0.7 min$
$t = 0.7 \times 60$
$t = 42 second$

AMU-2012

Heat Transfer

149650 A body cools from $60^{\circ} C$ to $50^{\circ} C$ in $10 \min$ . If room temperature is $25^{\circ} C$ of body at the end of next 10 min, will be

1

{38.5}^{\circ} C

2

40^{\circ} C

3

45^{\circ} C

4

{42.85}^{\circ} C

Explanation:

D Given,
Initial Temp $(θ_{1}) = 60^{\circ} C$ , Final Temp. $(θ_{2}) = 50^{\circ} C$
Atmospheric Temp. $(θ) = 25^{\circ} C$
Time $(t) = 10$ min
According to Newton's law,
$\frac{θ_{1} - θ_{2}}{t} = - K (\frac{θ_{1} + θ_{2}}{2} - θ)$
$\frac{60 - 50}{10} = - K (\frac{60 + 50}{2} - 25)$
$1 = - K (55 - 25)$
$K = - \frac{1}{30}$
Case-II
Again using Newton's law cooling,
$\frac{50 - θ_{3}}{10} = - (- \frac{1}{30}) (\frac{50 + θ_{3}}{2} - 25)$
$150 - 3 θ_{3} = \frac{50 + θ_{3} - 50}{2}$
$300 - 6 θ_{3} = θ_{3}$
$7 θ_{3} = 300$
$θ_{3} = {42.85}^{\circ} C$

UP CPMT-2012

Heat Transfer

149647 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2 \min$ , when the room temperature is $20^{\circ} C$ . The time taken for the food to cool from $86^{\circ} C$ to $74^{\circ} C$ will be

1

500 s

2

420 s

3

200 s

4

210 s

Explanation:

D Given, $T_{1} = 94^{\circ} C, T_{2} = 86^{\circ} C, t = 2 \min$
According to Newton's law of cooling-
$\frac{T_{1} - T_{2}}{2} = K [\frac{T_{1} + T_{2}}{2} - T_{s}]$
Where, $T_{s}$ is the surrounding temperature which is $20^{\circ} C$ . For first case,
$\frac{94 - 86}{2 \times 60} = K [\frac{94 + 86}{2} - 20]$
$\frac{8}{2 \times 60} = K \times 70$
$\frac{1}{15} = K \times 70$
For second case,
$\frac{86 - 74}{t} = K [\frac{86 + 74}{2} - 20]$
$\frac{12}{t} = K \times 60$
Dividing equation (i) by equation (ii), we get
$\frac{\frac{1}{15}}{\frac{12}{t}} = \frac{K \times 70}{K \times 60}$
$\frac{t}{12 \times 15} = \frac{7}{6}$
$t = 30 \times 7$
$t = 210 s$

AIIMS-27.05.2018(M)]

Heat Transfer

149648 A pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2 \min$ . When the room temperature is $20^{\circ} C$ . How long will it cool from $74^{0} C$ to $66^{\circ} C$ ?

1

2 \min

2

2.8 \min

3

2.5 \min

4

1.8 \min

Explanation:

B According to Newton's law of cooling
$\frac{θ_{1} - θ_{2}}{t} = K [\frac{θ_{1} + θ_{2}}{2} - θ_{o}]$
For case - (I)
$t = 2 \min$
$θ_{1} - θ_{2} = 94 - 86 = 8^{\circ} C$
$θ_{o} = 20^{\circ} C$
$\frac{θ_{1} + θ_{2}}{2} = \frac{94 + 86}{2} = \frac{180}{2} = 90^{\circ} C$
So, put these value in Newton's law of cooling formula we get
$\frac{8}{2} = K (90 - 20)$
$K = \frac{4}{70}$
For case - (II)
$θ_{1} - θ_{2} = 74 - 66 = 8^{\circ} C$
$t =$ ?
$\frac{θ_{1} + θ_{2}}{2} = \frac{74 + 66}{2} = \frac{140}{2} = 70^{\circ} C$
$θ_{o} = 20^{\circ} C$
Thus, using Newton's cooling law,
$\frac{8}{t} = K (70^{\circ} - 20^{\circ})$ Put the value of $K$ we get,
$\frac{8}{t} = \frac{4}{70} (50^{\circ})$
$t = \frac{70}{25} = 2.8$
Thus, it will take $2.8 \min$ .

Karnataka CET-2016

Heat Transfer

149649 Hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes when the room temperature is $20^{\circ} C$ . How long would the food take to cool from $71^{\circ} C$ to $69^{\circ} C$ ?

1

12 \sec

2

25 \sec

3

16 \sec

4

42 \sec

Explanation:

D We know Newton's law of cooling -
$\frac{T_{1} - T_{2}}{t} = - K (\frac{T_{1} + T_{2}}{2} - T_{S})$
For case-I
$T_{1} = 94^{\circ} C, T_{2} = 86^{\circ} C, t = 2 \min$ and $T_{s} = 20^{\circ} C$
$\frac{94 - 86}{2} = - K (\frac{94 + 86}{2} - 20)$
$K = - (4 / 70) = - (2 / 35)$
For case - II
$T_{1} = 71^{\circ} C, T_{2} = 69^{\circ} C, T_{s} = 20^{\circ} C$ and $t =$ ?
From equation (i)
$\frac{71 - 69}{t} = - (- \frac{2}{35}) \times (\frac{71 + 69}{2} - 20)$
$t = 0.7 min$
$t = 0.7 \times 60$
$t = 42 second$

AMU-2012

Heat Transfer

149650 A body cools from $60^{\circ} C$ to $50^{\circ} C$ in $10 \min$ . If room temperature is $25^{\circ} C$ of body at the end of next 10 min, will be

1

{38.5}^{\circ} C

2

40^{\circ} C

3

45^{\circ} C

4

{42.85}^{\circ} C

Explanation:

D Given,
Initial Temp $(θ_{1}) = 60^{\circ} C$ , Final Temp. $(θ_{2}) = 50^{\circ} C$
Atmospheric Temp. $(θ) = 25^{\circ} C$
Time $(t) = 10$ min
According to Newton's law,
$\frac{θ_{1} - θ_{2}}{t} = - K (\frac{θ_{1} + θ_{2}}{2} - θ)$
$\frac{60 - 50}{10} = - K (\frac{60 + 50}{2} - 25)$
$1 = - K (55 - 25)$
$K = - \frac{1}{30}$
Case-II
Again using Newton's law cooling,
$\frac{50 - θ_{3}}{10} = - (- \frac{1}{30}) (\frac{50 + θ_{3}}{2} - 25)$
$150 - 3 θ_{3} = \frac{50 + θ_{3} - 50}{2}$
$300 - 6 θ_{3} = θ_{3}$
$7 θ_{3} = 300$
$θ_{3} = {42.85}^{\circ} C$

UP CPMT-2012

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