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GRE Subject Test: Chemistry : Heat and Temperature

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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 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.

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

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

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

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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 : Heat and Temperature

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

All GRE Subject Test: Chemistry Resources

Example Questions

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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