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GRE Subject Test: Chemistry : Thermodynamics and Phases

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Example Question #361 :Gre Subject Test: Chemistry

Calculate the final temperature when a 44.0g sample of metal (specific heat of the metal= $0.52 \frac{J}{g^{o}C}$ ) at $45.0^{o}C$ is placed into 95.0g of water at $23.0^{o}C$ ．(specific heat of water is $4.18\frac{J}{g^{o}C}$ )

Possible Answers:

$76^{o}C$

$24^{o}C$

$132^{o}C$

$17^{o}C$

Correct answer:

$24^{o}C$

Explanation:

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1^oC. 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.

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

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

Heat is transferred from the metal to the water.

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

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

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

$-(44.0g\cdot 0.52\frac{J}{g^{o}C}\cdot (T_{f}-45.0^oC)=95.0g\cdot4.18\frac{J}{g^o}C(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}(T_{f}-45.0^{o}C))=397\frac{J}{^{o}C}(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-22.9\frac{J}{^{o}C}\cdot 45.0^{o}C)=397\frac{J}{^{o}C}\cdot T_{f}-397\frac{J}{^{o}C}\cdot 23.0^{o}C$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-1031J)=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}+1031J=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}-397\frac{J}{^{o}C}\cdot T_{f}=-1031J-9131J$

$-419\frac{J}{^{o}C}\cdot T_{f}=-10162J$

$T_{f}=\frac{-10162^{o}C}{-419.9}=24^{o}C$

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Example Question #1 :Heat And Temperature

Calculate the specific heat of 19.3 g of metal that requires 353 J of heat energy to raise its temperature from $17.0^{o}C$ to $44.0^{o}C$ ?

Possible Answers:

$0.98\frac{J}{g^{o}C}$

$0.677\frac{J}{g^{o}C}$

$0.581\frac{J}{g^{o}C}$

$17\frac{J}{g^{o}C}$

Correct answer:

$0.677\frac{J}{g^{o}C}$

Explanation:

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

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

$C=\frac{Q}{m\bigtriangleup T}=\frac{353J}{(19.3g\times (44.0^{o}C-17.0^{o}C))}=0.677\frac{J}{g^{o}C}$

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Example Question #1 :Heat And Temperature

What was the final temperature of the water if a 33.0 g sample of water absorbs 423 J of heat energy and heats up from $23^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

Possible Answers:

$54^{o}C$

$117^{o}C$

$71^{o}C$

$28^{o}C$

Correct answer:

$28^{o}C$

Explanation:

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

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

Rearranging gives,

$Q=mC\bigtriangleup T$

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

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

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

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

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

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

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Example Question #1 :Heat And Temperature

A 20.5 g metal at $37.1^{o}C$ was placed in 23.3 g of water at $24.2^{o}C$ ．The final temp of the water and metal was $26.0^{o}C$ ．Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ ．

Possible Answers:

$1.2\frac{J}{g^{o}C}$

$2.32\frac{J}{g^{o}C}$

$0.897\frac{J}{g^{o}C}$

$0.77\frac{J}{g^{o}C}$

Correct answer:

$0.77\frac{J}{g^{o}C}$

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.

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

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

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

$-(20.5g\cdot C\cdot (26.0^{o}C-37.1^oC)=23.3g\cdot4.18\frac{J}{g^o}C(26.0^{o}C-24.2^{o}C)$

$-(-227.55g^{o}C\cdot C)=175.3092J$

$227.55g^{o}C\cdot\ C=175.3092J$

$C=\frac{175.3092J}{227.55g^{o}C}=0.77\frac{J}{g^{o}C}$

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Example Question #41 :Physical Chemistry

What was the final temperature of the water if a 112 g sample of water absorbs 231 J of heat energy and heats up from $25.0^{o}C$ ? (specific heat of water= $4.18\frac{J}{g^{o}C}$ )

Possible Answers:

$43.2^{o}C$

$75^{o}C$

$25.5^{o}C$

$134^{o}C$

Correct answer:

$25.5^{o}C$

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,

$Q=mC\bigtriangleup T$

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

$231J=112g\times4.18\frac{J}{g^{o}C}\times(T_{f}-25.0^{o}C)$

$231J=468.16\frac{J}{^{o}C}\times(T_{f}-25.0^{o}C)$

$\frac{231J}{468.16\frac{J}{^{o}C}}=T_{f}-25.0^{o}C$

$0.493^{o}C=T_{f}-25.0^{o}C$

$T_{f}=25.5^{o}C$

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Example Question #1 :Heat And Temperature

A 10.2 g metal at $72.9^{o}C$ was placed in 21.2 g of water at $25.0^{o}C$ ．The final temp of the water and metal was $53.3^{o}C$ ．Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ ．

Possible Answers:

$2.6\frac{J}{g^{o}C}$

$12.5\frac{J}{g^{o}C}$

$465\frac{J}{g^{o}C}$

$173\frac{J}{g^{o}C}$

Correct answer:

$12.5\frac{J}{g^{o}C}$

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.

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

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

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

$-(10.2g\cdot C\cdot (53.3^{o}C-72.9^oC)=21.2g\cdot4.18\frac{J}{g^o}C(53.3^{o}C-25.0^{o}C)$

$-(-199.92g^{o}C\cdot C)=2507.83J$

$199.92g^{o}C\cdot C=2507.83J$

$C=\frac{2507.83J}{199.92g^{o}C}=12.5\frac{J}{g^{o}C}$

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Example Question #1 :Heat And Temperature

How muchheatis absorbed when 150 g of water goes from $25.0^{o}C$ to $35.0^{o}C$ ? (Specific heat of water= $4.18 \frac{J}{g^{o}C}$ )

Possible Answers:

$2270\ J$

$12\ J$

$70\ J$

$6270\ J$

Correct answer:

$6270\ J$

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.

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

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

$Q=mC\bigtriangleup T$

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

$Q=150g\cdot\ 4.18\frac{J}{g^{o}C}\cdot (35^{o}C-25^{o}C)$

$Q=6270\ J$

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Example Question #11 :General Thermodynamics

Calculate the specific heat of 20.4 g of metal that requires 410.4 J of heat energy to raise its temperature from $20.46^{o}C$ to $64.4^{o}C$ ?

Possible Answers:

$1.6\frac{J}{g^{o}C}$

$0.21\frac{J}{g^{o}C}$

$0.516\frac{J}{g^{o}C}$

$4.2\frac{J}{g^{o}C}$

Correct answer:

$0.516\frac{J}{g^{o}C}$

Explanation:

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

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

$C=\frac{Q}{m\bigtriangleup T}=\frac{410.4J}{(20.4g\times (64.4^{o}C-25.4^{o}C))}=0.516\frac{J}{g^{o}C}$

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Example Question #31 :Thermodynamics And Phases

Calculate the final temperature when a 14.7g sample of metal (specific heat of the metal= $0.72 \frac{J}{g^{o}C}$ ) at $37.0^{o}C$ is placed into 31.7g of water at $26.2^{o}C$ ．(Specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

Possible Answers:

54^oC

18^oC

36^oC

27^oC

Correct answer:

27^oC

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.

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

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

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

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

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

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

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

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

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

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

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

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Example Question #362 :Gre Subject Test: Chemistry

How muchheatis absorbed when 50 g of water goes from $24.5^{o}C$ to $43.0^{o}C$ ? (Specific heat of water = $4.18 \frac{J}{g^{o}C}$ )

Possible Answers:

896J

3867J

166J

43J

Correct answer:

3867J

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,

$Q=mC\bigtriangleup T$

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

$Q=50g\cdot\ 4.18\frac{J}{g^{o}C}\cdot (43.0^{o}C-24.5^{o}C)$

$Q=3867\ J$

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GRE Subject Test: Chemistry : Thermodynamics and Phases

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #361 :Gre Subject Test: Chemistry

Example Question #1 :Heat And Temperature

Example Question #1 :Heat And Temperature

Example Question #1 :Heat And Temperature

Example Question #41 :Physical Chemistry

Example Question #1 :Heat And Temperature

Example Question #1 :Heat And Temperature

Example Question #11 :General Thermodynamics

Example Question #31 :Thermodynamics And Phases

Example Question #362 :Gre Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

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