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Specific Heat

 

Specific Heat Capacity (C or S ) - The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity of the substance. The quantity of heat is frequently measured in units of Joules(J). Another property, the specific heat, is the heat capacity of the substance per gram of the substance. The specific heat of water is 4.18 J/g° C.

 

Substance  C (J/g oC)

 Air

  1.01
 Aluminum  0.902
 Copper  0.385
 Gold  0.129
 Iron  0.450
 Mercury  0.140
 NaCl  0.864
 Ice  2.03
 Water  4.18
   

q = m x C x DT

q = m x C x (Tf - Ti)

q = amount of heat energy gained or lost by substance

m = mass of sample

C = heat capacity (J oC-1 g-1 or J K-1 g-1)
Tf = final temperature
Ti = initial temperature

                                                          Highlight Answer Box to reveal answers

Sample Questions Answers
1. Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 46oC. q = m x C x DT

q = 250g x 4.18J/goC x 26oC 

q = 37,620J or 38kJ
2. Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25o to 60o. q = m x C x DT

C= q/m x DT

C = 204.75J /(15g x 35oC )

C= 0.39 J/goC
3. 216 J of energy is required to raise the temperature of aluminum from 15o to 35oC. Calculate the mass of aluminum.
(Specific Heat Capacity of aluminum is 0.90 JoC-1g-1).		 q = m x C x DT

m= q/C  x DT

m= 216J/(0.90J/goC x 20oC )

m= 12g
4. The initial temperature of 150g of ethanol was 22oC. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol?
(Specific heat capacity of ethanol is 2.44 JoC-1g-1).		 q = m x C x DT

DT =  q/m x C

DT = 3240J/(150g x 2.44J/goC)

DT = 8.85oC

Tfinal= 22oC +8.85oC= 30.9oC

More Practice Questions

Question Highlight to reveal answers
1. The temperature of a piece of Metal X with a mass of 95.4g increases from 25.0°C to 48.0°C as the metal absorbs 849 J of heat. What is the specific heat of Metal X?  Answer: 849 J /(95.4g x 23.0°C)

0.387 J/g°C

 
2. When 435 J of heat is added to 3.4 g of olive oil at 21°C, the temperature increases to 85°C. What is the specific heat of the olive oil?  Answer: 435 J/(3.4g x 64°C)

2.0 J/g°C
3. A piece of stainless steel with a mass of 1.55 g absorbs 141 J of heat when its temperature increases by 178°C. What is the specific heat of the stainless steel?  Answer: 141 J/(1.55 g x 178°C)

0.511 J/g°C
4. How much heat is required to raise the temperature of 250.0 g of mercury by 52°C?  Answer: 250.0 g x 0.140 J/g°C x 52°C

1800 J
6. How many kilojoules of heat are absorbed when 1.00 L of water is heated from 18°C to 85°C?  (Hint: You first need to determine the mass of the water, then calculate q in the requested unit.) Answer: 1.00kg x 4.18  J/g°C x 67°C

 280 kJ
7. A piece of aluminum with a mass of 100.0 g has a temperature of 20.0°C. It absorbs 1100 J of heat energy. What is the final temperature of the metal?  Answer:1100 J/(100.0 g x 0.902J/g°C)=120.°C + 20°C= 140.0°C
8. An unknown metal has a mass of 18.0 g. If the temperature of the metal sample rises from 15.0°C to 40.0°C as the sample absorbs 89.0 J of heat, what is the specific heat of the sample? Now look at your periodic table and choose a metal that is most likely the identity of the sample.  Answer: 89.0 J/(18.0 g x 25.0°C)

specific heat = 0.371 J/g°C
What is the final temperature after a 21.5 gram piece of ice at 0 is placed into a styrofoam cup with 125.0 grams of water initially at 76.5? Assume no loss or gain of heat from the surroundings.

Highlight Answer Below

energy to melt the ice

q=mHf
q=21.5 x 334j/g=7181J

q ice=q water (negative heat the water will cool down)
-7181J= 125.0g(4.18j/gC)(Tf-76.5C)
-13.7C=Tf-76.5
62.8C

A 24.57g sample of metal at 75.2 degrees Celcius was placed in 115.43g water at 22.6 degrees Celcius. The final temperature of the water and metal was 34.6 Celcius. If no heat was lost to the surroundings what is the specific heat of the metal?

Highlight Answer Below

qwater=qmetal
mc D T=mc DT
mc(Tf-Ti)=mc(Tf-Ti)
115.43g(4.18J/gC)(34.6C-22.6C)=24.57g x C x(34.6C-75.2C)

5760J=-997.5goC xC

5.77J/goC =C

 

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