GRE Subject Test: Chemistry - Physical Chemistry

What was the final temperature of the water if a sample of water absorbs of heat energy and heats up from $23^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

$28^{o}C$

$71^{o}C$

$117^{o}C$

$54^{o}C$

Explanation

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$423J=19g\times4.18\frac{J}{g^{o}C}\times(T_{f}-23^{o}C)$

$423J=79\frac{J}{^{o}C}\times(T_{f}-23^{o}C)$

$\frac{423J}{79\frac{J}{^{o}C}}=T_{f}-23^{o}C$

$5.35^{o}C=T_{f}-23^{o}C$

$T_{f}=28^{o}C$

Calculate the final temperature when a sample of metal (specific heat of the metal= $0.72 \frac{J}{g^{o}C}$ ) at $37.0^{o}C$ is placed into of water at $26.2^{o}C$ . (Specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

Explanation

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(14.7g\cdot 0.72\frac{J}{g^{o}C}\cdot (T_{f}-37.0^oC)=31.7g\cdot4.18\frac{J}{g^o}C(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}(T_{f}-37.0^{o}C))=132.51\frac{J}{^{o}C}(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-10.584\frac{J}{^{o}C}\cdot 37.0^{o}C)=132.51\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot 26.2^{o}C$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-391.61J)=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}+391.61J=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot T_{f}=-3472J-391.61J$

$-143.094\frac{J}{^{o}C}\cdot T_{f}=-3863.61J$

$T_{f}=\frac{-3863.61^{o}C}{-143.094}=27^{o}C$

Explanation

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(14.7g\cdot 0.72\frac{J}{g^{o}C}\cdot (T_{f}-37.0^oC)=31.7g\cdot4.18\frac{J}{g^o}C(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}(T_{f}-37.0^{o}C))=132.51\frac{J}{^{o}C}(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-10.584\frac{J}{^{o}C}\cdot 37.0^{o}C)=132.51\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot 26.2^{o}C$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-391.61J)=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}+391.61J=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot T_{f}=-3472J-391.61J$

$-143.094\frac{J}{^{o}C}\cdot T_{f}=-3863.61J$

$T_{f}=\frac{-3863.61^{o}C}{-143.094}=27^{o}C$

What was the final temperature of the water if a sample of water absorbs of heat energy and heats up from $23^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

$28^{o}C$

$71^{o}C$

$117^{o}C$

$54^{o}C$

Explanation

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$423J=19g\times4.18\frac{J}{g^{o}C}\times(T_{f}-23^{o}C)$

$423J=79\frac{J}{^{o}C}\times(T_{f}-23^{o}C)$

$\frac{423J}{79\frac{J}{^{o}C}}=T_{f}-23^{o}C$

$5.35^{o}C=T_{f}-23^{o}C$

$T_{f}=28^{o}C$

Explanation

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(14.7g\cdot 0.72\frac{J}{g^{o}C}\cdot (T_{f}-37.0^oC)=31.7g\cdot4.18\frac{J}{g^o}C(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}(T_{f}-37.0^{o}C))=132.51\frac{J}{^{o}C}(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-10.584\frac{J}{^{o}C}\cdot 37.0^{o}C)=132.51\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot 26.2^{o}C$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-391.61J)=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}+391.61J=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot T_{f}=-3472J-391.61J$

$-143.094\frac{J}{^{o}C}\cdot T_{f}=-3863.61J$

$T_{f}=\frac{-3863.61^{o}C}{-143.094}=27^{o}C$

What was the final temperature of the water if a sample of water absorbs of heat energy and heats up from $23^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

$28^{o}C$

$71^{o}C$

$117^{o}C$

$54^{o}C$

Explanation

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$423J=19g\times4.18\frac{J}{g^{o}C}\times(T_{f}-23^{o}C)$

$423J=79\frac{J}{^{o}C}\times(T_{f}-23^{o}C)$

$\frac{423J}{79\frac{J}{^{o}C}}=T_{f}-23^{o}C$

$5.35^{o}C=T_{f}-23^{o}C$

$T_{f}=28^{o}C$

What was the final temperature of the water if a sample of water absorbs of heat energy and heats up from $23^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

$28^{o}C$

$71^{o}C$

$117^{o}C$

$54^{o}C$

Explanation

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$423J=19g\times4.18\frac{J}{g^{o}C}\times(T_{f}-23^{o}C)$

$423J=79\frac{J}{^{o}C}\times(T_{f}-23^{o}C)$

$\frac{423J}{79\frac{J}{^{o}C}}=T_{f}-23^{o}C$

$5.35^{o}C=T_{f}-23^{o}C$

$T_{f}=28^{o}C$

Explanation

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(14.7g\cdot 0.72\frac{J}{g^{o}C}\cdot (T_{f}-37.0^oC)=31.7g\cdot4.18\frac{J}{g^o}C(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}(T_{f}-37.0^{o}C))=132.51\frac{J}{^{o}C}(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-10.584\frac{J}{^{o}C}\cdot 37.0^{o}C)=132.51\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot 26.2^{o}C$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-391.61J)=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}+391.61J=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot T_{f}=-3472J-391.61J$

$-143.094\frac{J}{^{o}C}\cdot T_{f}=-3863.61J$

$T_{f}=\frac{-3863.61^{o}C}{-143.094}=27^{o}C$

Two moles of sodium chloride (NaCl) are added to 1kg of a mystery solvent. The addition of the NaCl caused an increase of 6K to the solvent's boiling point.

Based on this information, what is the boiling constant for the solvent?

$k_{b} = 1.5\frac{K}{m}$

$k_{b} = 3.0\frac{K}{m}$

$k_{b}= 6.0\frac{K}{m}$

$k_{b}= 1.0\frac{K}{m}$

Explanation

In order to solve this problem, we can use the boiling point elevation equation: $\Delta T = k_{b}mi$ .

We know the temperature change, we can compute molality from the given information, and we know the van't Hoff factor (expected to be 2 in this scenario due to NaCl becoming 2 ions in solution). We can calculate the boiling point constant for the solvent.

$k_{b}= \frac{\Delta T}{mi}$

$m=\frac{mol}{kg}$

$k_{b} = \frac{6K}{(\frac{2moles}{1kg})(2)}$

$k_{b} = 1.5\frac{K}{m}$

Two moles of sodium chloride (NaCl) are added to 1kg of a mystery solvent. The addition of the NaCl caused an increase of 6K to the solvent's boiling point.

Based on this information, what is the boiling constant for the solvent?

$k_{b} = 1.5\frac{K}{m}$

$k_{b} = 3.0\frac{K}{m}$

$k_{b}= 6.0\frac{K}{m}$

$k_{b}= 1.0\frac{K}{m}$

Explanation

In order to solve this problem, we can use the boiling point elevation equation: $\Delta T = k_{b}mi$ .

$k_{b}= \frac{\Delta T}{mi}$

$m=\frac{mol}{kg}$

$k_{b} = \frac{6K}{(\frac{2moles}{1kg})(2)}$

$k_{b} = 1.5\frac{K}{m}$

Physical Chemistry

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