148189
$100 \mathrm{~g}$ ice is mixed with $100 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$. What will be the final temperature of the mixture?
1 $10^{\circ} \mathrm{C}$
2 $27^{\circ} \mathrm{C}$
3 $14^{\circ} \mathrm{C}$
4 none of these
Explanation:
A Heat given by water, $\mathrm{Q}_{\text {water }} =\mathrm{ms} \Delta \mathrm{T}$ $= 100 \times 1 \times(100-\mathrm{T})$ $=100 \times(100-\mathrm{T}) \mathrm{cal}$ Heat taken by ice, $\mathrm{Q}_{\text {ice }}=\mathrm{mL}+\mathrm{ms} \Delta \mathrm{T}$ $=100 \times 80+100 \times 1(\mathrm{~T}-0)$ $=8000+100 \mathrm{~T}$ According to principal of calorimetric, Heat given $=$ Heat taken $100 \times(100-T)=8000+100 T$ $10000-100 T=8000+100 \mathrm{~T}$ $2000=200 \mathrm{~T}$ $\mathrm{~T}=\frac{2000}{200}$ $=10^{\circ} \mathrm{C}$
JCECE-2003
Thermodynamics
148190
The latent heat of vaporisation of water is 2250 $\mathrm{J} / \mathrm{kg}$. If the work done in the process of vaporisation of $1 \mathrm{~kg}$ is $168 \mathrm{~J}$, then increase in internal energy will be:
148191
Assertion: In an isolated system the entropy increases. Reason: The processes in an isolated system are adiabatic.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B In non-isolated system, heat may enter into or escape from the system due to which entropy may increase or decrease but for isolated system we do not consider exchange of heat so, in this case entropy will always increase as the process is spontaneous. An adiabatic process involves no exchange of heat. We also define isolated system as having no exchange of heat with the surrounding so process in an isolated system are adiabatic. The two statements are independently correct but not co-related.
AIIMS-2006
Thermodynamics
148192
$\mathrm{N}_{2}$ gas is heated from $300 \mathrm{~K}$ temperature to $600 \mathrm{~K}$ through as isobaric process. Then find the change in the entropy of the gas. (Taken $n=$ 1 mole)
148189
$100 \mathrm{~g}$ ice is mixed with $100 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$. What will be the final temperature of the mixture?
1 $10^{\circ} \mathrm{C}$
2 $27^{\circ} \mathrm{C}$
3 $14^{\circ} \mathrm{C}$
4 none of these
Explanation:
A Heat given by water, $\mathrm{Q}_{\text {water }} =\mathrm{ms} \Delta \mathrm{T}$ $= 100 \times 1 \times(100-\mathrm{T})$ $=100 \times(100-\mathrm{T}) \mathrm{cal}$ Heat taken by ice, $\mathrm{Q}_{\text {ice }}=\mathrm{mL}+\mathrm{ms} \Delta \mathrm{T}$ $=100 \times 80+100 \times 1(\mathrm{~T}-0)$ $=8000+100 \mathrm{~T}$ According to principal of calorimetric, Heat given $=$ Heat taken $100 \times(100-T)=8000+100 T$ $10000-100 T=8000+100 \mathrm{~T}$ $2000=200 \mathrm{~T}$ $\mathrm{~T}=\frac{2000}{200}$ $=10^{\circ} \mathrm{C}$
JCECE-2003
Thermodynamics
148190
The latent heat of vaporisation of water is 2250 $\mathrm{J} / \mathrm{kg}$. If the work done in the process of vaporisation of $1 \mathrm{~kg}$ is $168 \mathrm{~J}$, then increase in internal energy will be:
148191
Assertion: In an isolated system the entropy increases. Reason: The processes in an isolated system are adiabatic.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B In non-isolated system, heat may enter into or escape from the system due to which entropy may increase or decrease but for isolated system we do not consider exchange of heat so, in this case entropy will always increase as the process is spontaneous. An adiabatic process involves no exchange of heat. We also define isolated system as having no exchange of heat with the surrounding so process in an isolated system are adiabatic. The two statements are independently correct but not co-related.
AIIMS-2006
Thermodynamics
148192
$\mathrm{N}_{2}$ gas is heated from $300 \mathrm{~K}$ temperature to $600 \mathrm{~K}$ through as isobaric process. Then find the change in the entropy of the gas. (Taken $n=$ 1 mole)
148189
$100 \mathrm{~g}$ ice is mixed with $100 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$. What will be the final temperature of the mixture?
1 $10^{\circ} \mathrm{C}$
2 $27^{\circ} \mathrm{C}$
3 $14^{\circ} \mathrm{C}$
4 none of these
Explanation:
A Heat given by water, $\mathrm{Q}_{\text {water }} =\mathrm{ms} \Delta \mathrm{T}$ $= 100 \times 1 \times(100-\mathrm{T})$ $=100 \times(100-\mathrm{T}) \mathrm{cal}$ Heat taken by ice, $\mathrm{Q}_{\text {ice }}=\mathrm{mL}+\mathrm{ms} \Delta \mathrm{T}$ $=100 \times 80+100 \times 1(\mathrm{~T}-0)$ $=8000+100 \mathrm{~T}$ According to principal of calorimetric, Heat given $=$ Heat taken $100 \times(100-T)=8000+100 T$ $10000-100 T=8000+100 \mathrm{~T}$ $2000=200 \mathrm{~T}$ $\mathrm{~T}=\frac{2000}{200}$ $=10^{\circ} \mathrm{C}$
JCECE-2003
Thermodynamics
148190
The latent heat of vaporisation of water is 2250 $\mathrm{J} / \mathrm{kg}$. If the work done in the process of vaporisation of $1 \mathrm{~kg}$ is $168 \mathrm{~J}$, then increase in internal energy will be:
148191
Assertion: In an isolated system the entropy increases. Reason: The processes in an isolated system are adiabatic.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B In non-isolated system, heat may enter into or escape from the system due to which entropy may increase or decrease but for isolated system we do not consider exchange of heat so, in this case entropy will always increase as the process is spontaneous. An adiabatic process involves no exchange of heat. We also define isolated system as having no exchange of heat with the surrounding so process in an isolated system are adiabatic. The two statements are independently correct but not co-related.
AIIMS-2006
Thermodynamics
148192
$\mathrm{N}_{2}$ gas is heated from $300 \mathrm{~K}$ temperature to $600 \mathrm{~K}$ through as isobaric process. Then find the change in the entropy of the gas. (Taken $n=$ 1 mole)
148189
$100 \mathrm{~g}$ ice is mixed with $100 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$. What will be the final temperature of the mixture?
1 $10^{\circ} \mathrm{C}$
2 $27^{\circ} \mathrm{C}$
3 $14^{\circ} \mathrm{C}$
4 none of these
Explanation:
A Heat given by water, $\mathrm{Q}_{\text {water }} =\mathrm{ms} \Delta \mathrm{T}$ $= 100 \times 1 \times(100-\mathrm{T})$ $=100 \times(100-\mathrm{T}) \mathrm{cal}$ Heat taken by ice, $\mathrm{Q}_{\text {ice }}=\mathrm{mL}+\mathrm{ms} \Delta \mathrm{T}$ $=100 \times 80+100 \times 1(\mathrm{~T}-0)$ $=8000+100 \mathrm{~T}$ According to principal of calorimetric, Heat given $=$ Heat taken $100 \times(100-T)=8000+100 T$ $10000-100 T=8000+100 \mathrm{~T}$ $2000=200 \mathrm{~T}$ $\mathrm{~T}=\frac{2000}{200}$ $=10^{\circ} \mathrm{C}$
JCECE-2003
Thermodynamics
148190
The latent heat of vaporisation of water is 2250 $\mathrm{J} / \mathrm{kg}$. If the work done in the process of vaporisation of $1 \mathrm{~kg}$ is $168 \mathrm{~J}$, then increase in internal energy will be:
148191
Assertion: In an isolated system the entropy increases. Reason: The processes in an isolated system are adiabatic.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B In non-isolated system, heat may enter into or escape from the system due to which entropy may increase or decrease but for isolated system we do not consider exchange of heat so, in this case entropy will always increase as the process is spontaneous. An adiabatic process involves no exchange of heat. We also define isolated system as having no exchange of heat with the surrounding so process in an isolated system are adiabatic. The two statements are independently correct but not co-related.
AIIMS-2006
Thermodynamics
148192
$\mathrm{N}_{2}$ gas is heated from $300 \mathrm{~K}$ temperature to $600 \mathrm{~K}$ through as isobaric process. Then find the change in the entropy of the gas. (Taken $n=$ 1 mole)
