Heat and Temperature - GRE Subject Test: Chemistry

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Question

Use the following values for water as needed.

$\Delta H^o_{fus}=334\frac{J}{g}$

$\Delta H^o_{vap}=2260\frac{J}{g}$

$c=4.186\frac{J}{g^oC}$

$\rho = 1\frac{g}{mL}$

If burning wood releases $\small 20kJ$ of heat energy per gram of wood consumed, what mass of wood must be consumed to heat $\small 1L$ of water from $\small 20^oC$ to $\small 100^oC$ , and then to convert it to water vapor?

Answer

There are two processes requiring added heat in this problem:

1. Raising the temperature of the liquid water from $\small 20^oC$ to $\small 100^oC$ (use $\Delta Q = mc\Delta T$ )

2. Boiling the water at a constant temperature of $\small 100^oC$ (use $\Delta Q = m\Delta H^o_{vap}$ )

To use either of these equation, we need to find the mass of the water using the relation between mass, density, and volume.

$m = \rhoV= 1\frac{g}{mL}1000mL=1000g$

Use this mass with the given specific heat and temperatures to find the heat for part 1 of the process.

$\Delta Q = (1000g)(4.186\frac{J}{g^oC})(100^oC - 20^oC)=334,880J$

Then, use the mass with the given heat of vaporization to find the energy needed to convert the water to water vapor.

$\Delta Q =(1000g) (2260\frac{J}{g})= 2,260,000J$

Sum the energies for step 1 and step 2.

$334,880 J + 2,260,000 J = 2,594,880J\approx 2,595kJ$

This is the total amount of energy needed from the burning wood. Use stoichiometry to find the grams of wood needed to produce this amount of energy.

$2595kJ*(\frac{1 g}{20kJ})=129.7g\approx 130g$

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Question

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

Answer

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.

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

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

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

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Question

What was the final temperature of water if a sample of water absorbs of heat energy and heats up from $25^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

Answer

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.

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

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

$Q=mC\bigtriangleup T$

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

$810.5J=19g\times4.179\frac{J}{g^{o}C}\times(T_{f}-25^{o}C)$

$810.5J=79.40\frac{J}{^{o}C}\times(T_{f}-25^{o}C)$

$\frac{810.5J}{79.40\frac{J}{^{o}C}}=T_{f}-25^{o}C$

$10.21^{o}C=T_{f}-25^{o}C$

$T_{f}=35^{o}C$

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Question

An metal at $54.2^{o}C$ was put into of water at $25^{o}C$ . The final temp of the water and metal was $29.3^{o}C$ . Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ .

Answer

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.

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

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

Heat is transferred from the metal to the water.

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

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

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

$-(81.9g\cdot C\cdot (29.3^{o}C-54.2^oC)=40.7g\cdot4.18\frac{J}{g^o}C(29.3^{o}C-25^{o}C)$

$-(-2039g^{o}C\cdot C)=731J$

$2039g^{o}C\cdot\ C=731J$

$C=\frac{731J}{2039g^{o}C}=0.358\frac{J}{g^{o}C}$

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Question

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

Answer

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.

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

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

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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Question

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

Answer

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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Question

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

Answer

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.

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

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

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

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Question

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 $4.18 \frac{J}{g^{o}C}$ )

Answer

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,

$Q=mC\bigtriangleup T$

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

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

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

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

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

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

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Question

A metal at $37.1^{o}C$ was placed in 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 $4.18\frac{J}{g^{o}C}$ .

Answer

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.

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

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

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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Question

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 $4.18 \frac{J}{g^{o}C}$ )

Answer

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.

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

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

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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Question

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

Answer

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,

$Q=mC\bigtriangleup T$

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

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

$Q=3867\ J$

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Question

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

Answer

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,

$Q=mC\bigtriangleup T$

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

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

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

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

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

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

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Question

A metal at $72.9^{o}C$ was placed in 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 $4.18\frac{J}{g^{o}C}$ .

Answer

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.

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

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

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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