Question

Substance molar heat capacity (C.)/J•mol-1.°C-1 75.3 specific heat capacity (C.) /J•g-lo°C-1 0.384 H2O(1) Cu(s) C,H,OH(l) (ethanol) Fe(s) 111.5 0.449 1. Fill in the empty entries in the table above. 2. If the same amount of energy is transferred to 1.0 g samples of each of the substances listed above, order them from largest AT to smallest AT. Explain. 3. Which is the consequence of copper's relatively low specific heat (0.385 J/(g°C)) compared to water (4.18 J/(g°C)) on the temperature change when big amount of energy is transferred? 4. If 1.00 kJ of energy is transferred to a sample of 100.0 g of liquid water at 20.0°C, what will the final temperature of the water be? 5. If 1.00 kJ of energy is transferred to a sample of 100.0 g of solid copper at 20.0°C, what will the final temperature of the copper be? 35

Answer 1

Hi,

Hope you are doing well.

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PART 1:

For H₂O:

Given,

Molar heat capacity, $C_{m}=75.3\;J/mol.^{o}C$

Molar mass of water, $m=18\;g/mol$

Therefore, Specific heat capacity of water is given by,

$C_{p}=\frac{C_{m}}{m}=\frac{75.3\;J/mol.^{o}C}{18\;g/mol}$

$\mathbf{\therefore C_{p}=4.18\;J/g.^{o}C}$

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For copper:

Given,

Specific heat capacity, $C_{p}=0.384\;J/g.^{o}C$

Molar mass of copper, $m=63.5\;g/mol$

Therefore, Molar heat capacity of copper is given by,

$C_{m}=C_{p}\times m=0.384\;J/g.^{o}C\times63.5\;g/mol$

$\mathbf{\therefore C_{m}=24.4\;J/mol.^{o}C}$

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For ethanol:

Given,

Molar heat capacity, $C_{m}=111.5\;J/mol.^{o}C$

Molar mass of ethanol, $m=46\;g/mol$

Therefore, Specific heat capacity of ethanol is given by,

$C_{p}=\frac{C_{m}}{m}=\frac{111.5\;J/mol.^{o}C}{46\;g/mol}$

$\mathbf{\therefore C_{p}=2.424\;J/g.^{o}C}$

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For iron:

Given,

Specific heat capacity, $C_{p}=0.449\;J/g.^{o}C$

Molar mass of iron, $m=55.845\;g/mol$

Therefore, Molar heat capacity of iron is given by,

$C_{m}=C_{p}\times m=0.449\;J/g.^{o}C\times55.845\;g/mol$

$\mathbf{\therefore C_{m}=25.07\;J/mol.^{o}C}$

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TABLE:

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Substance molar heat capacity (C )/J.mol-1.°C-1 75.3 24.4 H2O(1) Cu(s) C,H,OH() (ethanol) Fe(s) specific heat capacity (C.) /Jog-1.°C-1 4.18 0.384 2.424 111.5 25.07 0.449

PART 2:

To answer the remaining questions, we should know the definition of specific heat capacity.

Specific heat capacity is defined as the amount of energy required to raise the temperature of 1 g substance through 1^oC.

$C=\frac{Heat}{mass\times\Delta T}$

Hence, When the heat supplied and amount of substances are same, then the substance with higher specific heat capacity will have lower temperature difference.

$C\propto \frac{1}{\Delta T}$

Hence the order of temperature difference is as follows (from largest $\Delta T$ to the smallest ) or (from smallest $C$ to the largest C).

1. Copper.
2. Iron.
3. Ethanol.
4. Water.

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PART 3:

As we said earlier, specific heat is inversely proportional to the temperature difference. Hence for copper (low C) the temperature difference will be very high as compared to water (high C).

Hence, for a large amount of energy temperature of copper will rise quickly as compared to water.

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PART 4:

Given that,

Energy supplied, $E=1\;kJ=1000\;J$

Mass of water, $m=100\;g$

Initial temperature, $T_{1}=20^{o}C$

Specific heat capacity of water, $C=4.18\;J/g^{o}C$

So final temperature is given by:

$H=mC(T_{2}-T_{1})$

$\therefore 1000\;J=100\;g\times4.18\;J/g^{o}C\times (T_{2}-20^{o}C)$

$\therefore T_{2}-20^{o}C=\frac{1000\;J}{100\;g\times4.18\;J/g^{o}C}$

$\therefore T_{2}-20^{o}C=2.39^{o}C$

$\mathbf{\therefore T_{2}=22.39^{o}C}$

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PART 5:

Given that,

Energy supplied, $E=1\;kJ=1000\;J$

Mass of copper, $m=100\;g$

Initial temperature, $T_{1}=20^{o}C$

Specific heat capacity of water, $C=0.384\;J/g^{o}C$

So final temperature is given by:

$H=mC(T_{2}-T_{1})$

$\therefore 1000\;J=100\;g\times0.384\;J/g^{o}C\times (T_{2}-20^{o}C)$

$\therefore T_{2}-20^{o}C=\frac{1000\;J}{100\;g\times0.384\;J/g^{o}C}$

$\therefore T_{2}-20^{o}C=26.04^{o}C$

$\mathbf{\therefore T_{2}=46.04^{o}C}$

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Hope this helped for your studies. Keep learning. Have a good day.

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