Question

just #6

Thermochemistry, Part I, Calorimetry, Homework 7 Specific Heat Values: cal; Diamond(0.124), graphite(0.170), ice(0.50), steam(0.48), Al(0.215), Fe(0.108); S(0.176); liquidCCIF(0.208), gasCCI;F(0.142), liquidCF4(0.294), gasCF4(0.169), Heat of Fusion values: cal; water(80.0), Al(94.4), Heat of Vaporization values: Car; water(540.0), CC13F(43.10), CF4(32.49), Al(1992) 1. Draw the heating-cooling curve for phenol which has a melting point of 43 ºC and a boiling point of 182°C. 2. Draw the heating-cooling curve for Aluminum which has a melting point of 660 °C and a boiling point of 1800 °C. Include the specific heat for the solid as well as the values for the heat of fusion and vaporization 3. How many J and cal are needed to heat 250 mg of diamond from -20 °C to 150 °C? 4. How many J and cal are needed to heat 250 mg of graphite from -20 °C to 150°C? 5. How many cal are needed to bring 1750 g ice from -20.0 °C to steam at 110 °C? 6. How many cal and J are needed to heat 34.7g Freon 11 (CC13F) from liquid at -11.5 °C to vapor at 50 °C?. Note, the melting point for CCI F is -111 °C and the boiling point is +23.8 °C.

Answer 1

Hi,

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SOLUTION:

PART 6:

This process will happen in 3 steps:

1. Raising the temperature of liquid Freon 11 (CCl₃F) from -11.5^oC to 23.8^oC (boiling point)
2. Phase change from liquid to vapor at 23.8^oC.
3. Raising the temperature of gaseous Freon 11 from 23.8^oC to 50^oC.

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STEP 1:Raising the temperature of liquid Freon 11 (CCl₃F) from -11.5^oC to 23.8^oC (boiling point):

Given that,

Mass of Freon 11 (CCl₃F), m = 34.7 g

Specific heat capacity of liquid Freon 11, $C=0.208\;cal/g^{o}C$

Change in temperature, $\Delta T=(23.8^{o}C-(-11.5)^{o}C)=35.3^{o}C$

Therefore heat required for this step is given by,

$H_{1}=mC\Delta T=34.7 \;g\times0.208\;cal/g^{o}C\times 35.3^{o}C$

$\boldsymbol{\therefore H_{1}=254.78128\;cal}$

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STEP 2:Phase change from liquid to vapor at 23.8^oC:

We have,

Mass of Freon 11 (CCl₃F), m = 34.7 g

Heat of vaporization Freon 11,

43. 10 сal/g Hуaр

Therefore heat required for this step is given by,

$H_{2}=mCH_{vap}=34.7 \;g\times43.10\;cal/g$

$\boldsymbol{\therefore H_{2}=1495.57\;cal}$

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STEP 3:Raising the temperature of gaseous Freon 11 from 23.8^oC to 50^oC.

Mass of Freon 11 (CCl₃F), m = 34.7 g

Specific heat capacity of gaseous Freon 11, $C=0.142\;cal/g^{o}C$

Change in temperature, $\Delta T=(50^{o}C-23.8^{o}C)=26.2^{o}C$

Therefore heat required for this step is given by,

$H_{3}=mC\Delta T=34.7 \;g\times0.142\;cal/g^{o}C\times 26.2^{o}C$

$\boldsymbol{\therefore H_{3}=129.09788\;cal}$

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Therefore total heat energy required for the entire process is given by,

$H=H_{1}+H_{2}+H_{3}$

$=254.78128\;cal+1495.57\;cal+129.09788\;cal$

$\boldsymbol{\therefore H=1879.45\;cal}$

We know that, 1 cal = 4.184 J.

$\therefore H=1879.45\;cal\times \frac{4.184\;J}{cal}$

$\boldsymbol{\therefore H=7863.62\;J}$

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