03. Newton's Law of Cooling and Seebeck Effect
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Heat Transfer

149612 A bowl filled with very hot soup cools from $98^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minute when the room temperature is $22^{\circ} \mathrm{C}$.
How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?

1 2 minute
2 1 minute
3 1.4 minute
4 0.5 minute
JEE Main-25.01.2023
Heat Transfer

149613 Hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the first $10 \mathrm{~min}$ and to $42^{\circ} \mathrm{C}$ in the next $10 \mathrm{~min}$. Then the temperature of the surrounding is:

1 $20^{\circ} \mathrm{C}$
2 $30^{\circ} \mathrm{C}$
3 $15^{\circ} \mathrm{C}$
4 $10^{\circ} \mathrm{C}$
Karnataka CET-2010
Heat Transfer

149614 The rate of loss of heat of a body is directly proportional to the difference of temperature of the body and the surroundings. This statement is known as

1 Stefan's Law
2 Newton's law of cooling
3 Wien's law
4 Kirchhoff's law
J and K CET- 2011
Heat Transfer

149616 A liquid cools from $70^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 5 minutes. Find the time taken by the liquid to cool from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$, if the temperature of the surrounding is constant at $30^{\circ} \mathrm{C}$.

1 3 minutes
2 5 minutes
3 7 minutes
4 9 minutes
J and K-CET-2019
Heat Transfer

149617 A body cools from $100^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in $8 \mathrm{~s}$. if the room temperature is $15^{\circ} \mathrm{C}$ and assuming Newton's law of cooling holds goods, then time required for the body to cool from $70^{0} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is

1 $14 \mathrm{~s}$
2 $8 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $5 \mathrm{~s}$
MHT-CET 2005
NEET DIGITAL LIBRARY
Heat Transfer

149612 A bowl filled with very hot soup cools from $98^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minute when the room temperature is $22^{\circ} \mathrm{C}$.
How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?

1 2 minute
2 1 minute
3 1.4 minute
4 0.5 minute
JEE Main-25.01.2023
Heat Transfer

149613 Hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the first $10 \mathrm{~min}$ and to $42^{\circ} \mathrm{C}$ in the next $10 \mathrm{~min}$. Then the temperature of the surrounding is:

1 $20^{\circ} \mathrm{C}$
2 $30^{\circ} \mathrm{C}$
3 $15^{\circ} \mathrm{C}$
4 $10^{\circ} \mathrm{C}$
Karnataka CET-2010
Heat Transfer

149614 The rate of loss of heat of a body is directly proportional to the difference of temperature of the body and the surroundings. This statement is known as

1 Stefan's Law
2 Newton's law of cooling
3 Wien's law
4 Kirchhoff's law
J and K CET- 2011
Heat Transfer

149616 A liquid cools from $70^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 5 minutes. Find the time taken by the liquid to cool from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$, if the temperature of the surrounding is constant at $30^{\circ} \mathrm{C}$.

1 3 minutes
2 5 minutes
3 7 minutes
4 9 minutes
J and K-CET-2019
Heat Transfer

149617 A body cools from $100^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in $8 \mathrm{~s}$. if the room temperature is $15^{\circ} \mathrm{C}$ and assuming Newton's law of cooling holds goods, then time required for the body to cool from $70^{0} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is

1 $14 \mathrm{~s}$
2 $8 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $5 \mathrm{~s}$
MHT-CET 2005
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Heat Transfer

149612 A bowl filled with very hot soup cools from $98^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minute when the room temperature is $22^{\circ} \mathrm{C}$.
How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?

1 2 minute
2 1 minute
3 1.4 minute
4 0.5 minute
JEE Main-25.01.2023
Heat Transfer

149613 Hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the first $10 \mathrm{~min}$ and to $42^{\circ} \mathrm{C}$ in the next $10 \mathrm{~min}$. Then the temperature of the surrounding is:

1 $20^{\circ} \mathrm{C}$
2 $30^{\circ} \mathrm{C}$
3 $15^{\circ} \mathrm{C}$
4 $10^{\circ} \mathrm{C}$
Karnataka CET-2010
Heat Transfer

149614 The rate of loss of heat of a body is directly proportional to the difference of temperature of the body and the surroundings. This statement is known as

1 Stefan's Law
2 Newton's law of cooling
3 Wien's law
4 Kirchhoff's law
J and K CET- 2011
Heat Transfer

149616 A liquid cools from $70^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 5 minutes. Find the time taken by the liquid to cool from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$, if the temperature of the surrounding is constant at $30^{\circ} \mathrm{C}$.

1 3 minutes
2 5 minutes
3 7 minutes
4 9 minutes
J and K-CET-2019
Heat Transfer

149617 A body cools from $100^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in $8 \mathrm{~s}$. if the room temperature is $15^{\circ} \mathrm{C}$ and assuming Newton's law of cooling holds goods, then time required for the body to cool from $70^{0} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is

1 $14 \mathrm{~s}$
2 $8 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $5 \mathrm{~s}$
MHT-CET 2005
For More Updates join with Us
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Heat Transfer

149612 A bowl filled with very hot soup cools from $98^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minute when the room temperature is $22^{\circ} \mathrm{C}$.
How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?

1 2 minute
2 1 minute
3 1.4 minute
4 0.5 minute
JEE Main-25.01.2023
Heat Transfer

149613 Hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the first $10 \mathrm{~min}$ and to $42^{\circ} \mathrm{C}$ in the next $10 \mathrm{~min}$. Then the temperature of the surrounding is:

1 $20^{\circ} \mathrm{C}$
2 $30^{\circ} \mathrm{C}$
3 $15^{\circ} \mathrm{C}$
4 $10^{\circ} \mathrm{C}$
Karnataka CET-2010
Heat Transfer

149614 The rate of loss of heat of a body is directly proportional to the difference of temperature of the body and the surroundings. This statement is known as

1 Stefan's Law
2 Newton's law of cooling
3 Wien's law
4 Kirchhoff's law
J and K CET- 2011
Heat Transfer

149616 A liquid cools from $70^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 5 minutes. Find the time taken by the liquid to cool from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$, if the temperature of the surrounding is constant at $30^{\circ} \mathrm{C}$.

1 3 minutes
2 5 minutes
3 7 minutes
4 9 minutes
J and K-CET-2019
Heat Transfer

149617 A body cools from $100^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in $8 \mathrm{~s}$. if the room temperature is $15^{\circ} \mathrm{C}$ and assuming Newton's law of cooling holds goods, then time required for the body to cool from $70^{0} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is

1 $14 \mathrm{~s}$
2 $8 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $5 \mathrm{~s}$
MHT-CET 2005
NEET DIGITAL LIBRARY
Heat Transfer

149612 A bowl filled with very hot soup cools from $98^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minute when the room temperature is $22^{\circ} \mathrm{C}$.
How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?

1 2 minute
2 1 minute
3 1.4 minute
4 0.5 minute
JEE Main-25.01.2023
Heat Transfer

149613 Hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the first $10 \mathrm{~min}$ and to $42^{\circ} \mathrm{C}$ in the next $10 \mathrm{~min}$. Then the temperature of the surrounding is:

1 $20^{\circ} \mathrm{C}$
2 $30^{\circ} \mathrm{C}$
3 $15^{\circ} \mathrm{C}$
4 $10^{\circ} \mathrm{C}$
Karnataka CET-2010
Heat Transfer

149614 The rate of loss of heat of a body is directly proportional to the difference of temperature of the body and the surroundings. This statement is known as

1 Stefan's Law
2 Newton's law of cooling
3 Wien's law
4 Kirchhoff's law
J and K CET- 2011
Heat Transfer

149616 A liquid cools from $70^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 5 minutes. Find the time taken by the liquid to cool from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$, if the temperature of the surrounding is constant at $30^{\circ} \mathrm{C}$.

1 3 minutes
2 5 minutes
3 7 minutes
4 9 minutes
J and K-CET-2019
Heat Transfer

149617 A body cools from $100^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in $8 \mathrm{~s}$. if the room temperature is $15^{\circ} \mathrm{C}$ and assuming Newton's law of cooling holds goods, then time required for the body to cool from $70^{0} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is

1 $14 \mathrm{~s}$
2 $8 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $5 \mathrm{~s}$
MHT-CET 2005