148701
The liquids at temperature $60^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$ respectively have masses in the ratio $3: 4$ and their specific heats in the ratio $4: 5$. If the two liquids are mixed, the resultant temperature is
1 $70^{\circ} \mathrm{C}$
2 $50^{\circ} \mathrm{C}$
3 $40^{\circ} \mathrm{C}$
4 $41.25^{\circ} \mathrm{C}$
Explanation:
D Given, $\mathrm{T}_{2}=60^{\circ} \mathrm{C}+273=333 \mathrm{~K}$ $\mathrm{~T}_{1}=30^{\circ} \mathrm{C}+273=303 \mathrm{~K}$ Mass ratio $\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)$ is $3: 4$ and specific heat ratio $\left(\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}\right)$ is $4: 5$. We know, $\mathrm{m}_{1} \mathrm{C}_{1}(60-\mathrm{T})=\mathrm{m}_{2} \mathrm{C}_{2}(\mathrm{~T}-30)$ $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \times \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $\frac{3}{4} \times \frac{4}{5}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $\frac{3}{5}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $180-3 \mathrm{~T}=5 \mathrm{~T}-150$ $8 \mathrm{~T}=330$ $\mathrm{~T}=41.25^{\circ} \mathrm{C}$
EAMCET-2000
Thermodynamics
148702
The latent heat of ice is $80 \mathrm{Cal} / \mathrm{gm}$. The change in entropy when 10 gram of ice at $0^{\circ} \mathrm{C}$ is converted into water of same temperature is
1 $0.293 \mathrm{Cal} / \mathrm{K}$
2 $2.93 \mathrm{Cal} / \mathrm{K}$
3 $80 \mathrm{Cal} / \mathrm{K}$
4 $8 \mathrm{Cal} / \mathrm{K}$
Explanation:
B Given, $\mathrm{L}_{\text {ice }}=80 \mathrm{cal} / \mathrm{gm}$ Mass of ice $=10 \mathrm{gm}$ Heat required to convert in water, $\mathrm{Q}=\mathrm{ml}$ $=10 \mathrm{gm} \times 80 \mathrm{cal} / \mathrm{gm}$. $=800 \mathrm{cal}$. $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}$ $\mathrm{S}=\frac{800}{273}$ $\mathrm{S}=2.93 \mathrm{Cal} / \mathrm{K}$
UPSEE 2019
Thermodynamics
148690
Even Carnot engine cannot give $100 \%$ efficiency because we cannot
1 prevent radiation
2 find ideal sources
3 reach absolute zero temperature
4 eliminate friction
Explanation:
C We know that the efficiency of a Carnot engine is given as: $\eta=1-\frac{T_{2}}{T_{1}}$ For $100 \%$ efficiency i.e., $\eta=1, \mathrm{~T}_{2}$ must be equal $0 \mathrm{~K} .0$ $\mathrm{K}$ is the absolute zero temperature i.e. the lowest limit of the thermodynamic temperature scale. Absolute zero indicates the temperature at which the particles in a matter are essentially motionless, which is physically impossible.
CG PET- 2006
Thermodynamics
148694
A measure of degree of disorder of a system is known as
1 isobaric
2 isotropy
3 entropy
4 enthalpy
Explanation:
C A measure of degree of disorderness of a system is known as entropy. The change in entropy is given as: $\Delta \mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}$ Where, $\Delta \mathrm{Q}=$ Heat absorbed by the system $\mathrm{T}=$ Absolute temperature
148701
The liquids at temperature $60^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$ respectively have masses in the ratio $3: 4$ and their specific heats in the ratio $4: 5$. If the two liquids are mixed, the resultant temperature is
1 $70^{\circ} \mathrm{C}$
2 $50^{\circ} \mathrm{C}$
3 $40^{\circ} \mathrm{C}$
4 $41.25^{\circ} \mathrm{C}$
Explanation:
D Given, $\mathrm{T}_{2}=60^{\circ} \mathrm{C}+273=333 \mathrm{~K}$ $\mathrm{~T}_{1}=30^{\circ} \mathrm{C}+273=303 \mathrm{~K}$ Mass ratio $\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)$ is $3: 4$ and specific heat ratio $\left(\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}\right)$ is $4: 5$. We know, $\mathrm{m}_{1} \mathrm{C}_{1}(60-\mathrm{T})=\mathrm{m}_{2} \mathrm{C}_{2}(\mathrm{~T}-30)$ $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \times \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $\frac{3}{4} \times \frac{4}{5}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $\frac{3}{5}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $180-3 \mathrm{~T}=5 \mathrm{~T}-150$ $8 \mathrm{~T}=330$ $\mathrm{~T}=41.25^{\circ} \mathrm{C}$
EAMCET-2000
Thermodynamics
148702
The latent heat of ice is $80 \mathrm{Cal} / \mathrm{gm}$. The change in entropy when 10 gram of ice at $0^{\circ} \mathrm{C}$ is converted into water of same temperature is
1 $0.293 \mathrm{Cal} / \mathrm{K}$
2 $2.93 \mathrm{Cal} / \mathrm{K}$
3 $80 \mathrm{Cal} / \mathrm{K}$
4 $8 \mathrm{Cal} / \mathrm{K}$
Explanation:
B Given, $\mathrm{L}_{\text {ice }}=80 \mathrm{cal} / \mathrm{gm}$ Mass of ice $=10 \mathrm{gm}$ Heat required to convert in water, $\mathrm{Q}=\mathrm{ml}$ $=10 \mathrm{gm} \times 80 \mathrm{cal} / \mathrm{gm}$. $=800 \mathrm{cal}$. $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}$ $\mathrm{S}=\frac{800}{273}$ $\mathrm{S}=2.93 \mathrm{Cal} / \mathrm{K}$
UPSEE 2019
Thermodynamics
148690
Even Carnot engine cannot give $100 \%$ efficiency because we cannot
1 prevent radiation
2 find ideal sources
3 reach absolute zero temperature
4 eliminate friction
Explanation:
C We know that the efficiency of a Carnot engine is given as: $\eta=1-\frac{T_{2}}{T_{1}}$ For $100 \%$ efficiency i.e., $\eta=1, \mathrm{~T}_{2}$ must be equal $0 \mathrm{~K} .0$ $\mathrm{K}$ is the absolute zero temperature i.e. the lowest limit of the thermodynamic temperature scale. Absolute zero indicates the temperature at which the particles in a matter are essentially motionless, which is physically impossible.
CG PET- 2006
Thermodynamics
148694
A measure of degree of disorder of a system is known as
1 isobaric
2 isotropy
3 entropy
4 enthalpy
Explanation:
C A measure of degree of disorderness of a system is known as entropy. The change in entropy is given as: $\Delta \mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}$ Where, $\Delta \mathrm{Q}=$ Heat absorbed by the system $\mathrm{T}=$ Absolute temperature
148701
The liquids at temperature $60^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$ respectively have masses in the ratio $3: 4$ and their specific heats in the ratio $4: 5$. If the two liquids are mixed, the resultant temperature is
1 $70^{\circ} \mathrm{C}$
2 $50^{\circ} \mathrm{C}$
3 $40^{\circ} \mathrm{C}$
4 $41.25^{\circ} \mathrm{C}$
Explanation:
D Given, $\mathrm{T}_{2}=60^{\circ} \mathrm{C}+273=333 \mathrm{~K}$ $\mathrm{~T}_{1}=30^{\circ} \mathrm{C}+273=303 \mathrm{~K}$ Mass ratio $\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)$ is $3: 4$ and specific heat ratio $\left(\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}\right)$ is $4: 5$. We know, $\mathrm{m}_{1} \mathrm{C}_{1}(60-\mathrm{T})=\mathrm{m}_{2} \mathrm{C}_{2}(\mathrm{~T}-30)$ $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \times \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $\frac{3}{4} \times \frac{4}{5}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $\frac{3}{5}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $180-3 \mathrm{~T}=5 \mathrm{~T}-150$ $8 \mathrm{~T}=330$ $\mathrm{~T}=41.25^{\circ} \mathrm{C}$
EAMCET-2000
Thermodynamics
148702
The latent heat of ice is $80 \mathrm{Cal} / \mathrm{gm}$. The change in entropy when 10 gram of ice at $0^{\circ} \mathrm{C}$ is converted into water of same temperature is
1 $0.293 \mathrm{Cal} / \mathrm{K}$
2 $2.93 \mathrm{Cal} / \mathrm{K}$
3 $80 \mathrm{Cal} / \mathrm{K}$
4 $8 \mathrm{Cal} / \mathrm{K}$
Explanation:
B Given, $\mathrm{L}_{\text {ice }}=80 \mathrm{cal} / \mathrm{gm}$ Mass of ice $=10 \mathrm{gm}$ Heat required to convert in water, $\mathrm{Q}=\mathrm{ml}$ $=10 \mathrm{gm} \times 80 \mathrm{cal} / \mathrm{gm}$. $=800 \mathrm{cal}$. $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}$ $\mathrm{S}=\frac{800}{273}$ $\mathrm{S}=2.93 \mathrm{Cal} / \mathrm{K}$
UPSEE 2019
Thermodynamics
148690
Even Carnot engine cannot give $100 \%$ efficiency because we cannot
1 prevent radiation
2 find ideal sources
3 reach absolute zero temperature
4 eliminate friction
Explanation:
C We know that the efficiency of a Carnot engine is given as: $\eta=1-\frac{T_{2}}{T_{1}}$ For $100 \%$ efficiency i.e., $\eta=1, \mathrm{~T}_{2}$ must be equal $0 \mathrm{~K} .0$ $\mathrm{K}$ is the absolute zero temperature i.e. the lowest limit of the thermodynamic temperature scale. Absolute zero indicates the temperature at which the particles in a matter are essentially motionless, which is physically impossible.
CG PET- 2006
Thermodynamics
148694
A measure of degree of disorder of a system is known as
1 isobaric
2 isotropy
3 entropy
4 enthalpy
Explanation:
C A measure of degree of disorderness of a system is known as entropy. The change in entropy is given as: $\Delta \mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}$ Where, $\Delta \mathrm{Q}=$ Heat absorbed by the system $\mathrm{T}=$ Absolute temperature
148701
The liquids at temperature $60^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$ respectively have masses in the ratio $3: 4$ and their specific heats in the ratio $4: 5$. If the two liquids are mixed, the resultant temperature is
1 $70^{\circ} \mathrm{C}$
2 $50^{\circ} \mathrm{C}$
3 $40^{\circ} \mathrm{C}$
4 $41.25^{\circ} \mathrm{C}$
Explanation:
D Given, $\mathrm{T}_{2}=60^{\circ} \mathrm{C}+273=333 \mathrm{~K}$ $\mathrm{~T}_{1}=30^{\circ} \mathrm{C}+273=303 \mathrm{~K}$ Mass ratio $\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)$ is $3: 4$ and specific heat ratio $\left(\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}\right)$ is $4: 5$. We know, $\mathrm{m}_{1} \mathrm{C}_{1}(60-\mathrm{T})=\mathrm{m}_{2} \mathrm{C}_{2}(\mathrm{~T}-30)$ $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \times \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $\frac{3}{4} \times \frac{4}{5}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $\frac{3}{5}=\frac{\mathrm{T}-30}{60-\mathrm{T}}$ $180-3 \mathrm{~T}=5 \mathrm{~T}-150$ $8 \mathrm{~T}=330$ $\mathrm{~T}=41.25^{\circ} \mathrm{C}$
EAMCET-2000
Thermodynamics
148702
The latent heat of ice is $80 \mathrm{Cal} / \mathrm{gm}$. The change in entropy when 10 gram of ice at $0^{\circ} \mathrm{C}$ is converted into water of same temperature is
1 $0.293 \mathrm{Cal} / \mathrm{K}$
2 $2.93 \mathrm{Cal} / \mathrm{K}$
3 $80 \mathrm{Cal} / \mathrm{K}$
4 $8 \mathrm{Cal} / \mathrm{K}$
Explanation:
B Given, $\mathrm{L}_{\text {ice }}=80 \mathrm{cal} / \mathrm{gm}$ Mass of ice $=10 \mathrm{gm}$ Heat required to convert in water, $\mathrm{Q}=\mathrm{ml}$ $=10 \mathrm{gm} \times 80 \mathrm{cal} / \mathrm{gm}$. $=800 \mathrm{cal}$. $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}$ $\mathrm{S}=\frac{800}{273}$ $\mathrm{S}=2.93 \mathrm{Cal} / \mathrm{K}$
UPSEE 2019
Thermodynamics
148690
Even Carnot engine cannot give $100 \%$ efficiency because we cannot
1 prevent radiation
2 find ideal sources
3 reach absolute zero temperature
4 eliminate friction
Explanation:
C We know that the efficiency of a Carnot engine is given as: $\eta=1-\frac{T_{2}}{T_{1}}$ For $100 \%$ efficiency i.e., $\eta=1, \mathrm{~T}_{2}$ must be equal $0 \mathrm{~K} .0$ $\mathrm{K}$ is the absolute zero temperature i.e. the lowest limit of the thermodynamic temperature scale. Absolute zero indicates the temperature at which the particles in a matter are essentially motionless, which is physically impossible.
CG PET- 2006
Thermodynamics
148694
A measure of degree of disorder of a system is known as
1 isobaric
2 isotropy
3 entropy
4 enthalpy
Explanation:
C A measure of degree of disorderness of a system is known as entropy. The change in entropy is given as: $\Delta \mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}$ Where, $\Delta \mathrm{Q}=$ Heat absorbed by the system $\mathrm{T}=$ Absolute temperature
