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This is a graph of temperature v. heat added for a .218 kg sample of unknown.  It starts as a solid, and ends as a gas.

 

1. What is the melting temperature and the boiling temperature of this substance?  Label the solid, liquid and gaseous phases.

(Melt = 30 ^oC, Boil = 50 ^oC, solid 0-300 J added, Liquid 700-1100 J added, Gas 1700-2000 J added.)

Where the temperature is rising, it is all one phase (i.e. solid, liquid, gas).  Where the temperature plateaus, the substance is undergoing a phase change - either melting, or boiling.

 

2. What is the latent heat of fusion?  (Melting)

(1830 J/kg)

Use

Q = 400 J (starts at 300 J, ends at 700 J)
m = 0.218 kg (see above the graph)
L = 1834.86 = 1830 J/kg

3. What is the specific heat of the liquid phase?

(91.7 J/kg^oC)

Use

Q = 400 J (starts at 700 J, ends at 1100 J - Read the graph)
m = 0.218 kg (see above the graph)
DT = 20 ^oC (starts at 30^oC ends at 50 ^oC - Just read the graph)
C = 91.743 = 91.7 J/kg^oC

 

4. What heat do you need to heat 3.29 Kg of water at 21.0 ^oC to steam at 175 ^oC?  (For H₂O: C_ice = 2100 J/kg^oC, l_f = 3.33 x 10⁵ J/Kg, C_water = 4186 J/Kg^oC, l_v = 22.6 x 10⁵ J/Kg, C_steam = 2010 J/kg^oC)

(9,019,350.76 J – 9.02 x 10⁶ J)

 

This will need to heat to boiling from 21.0 ^oC to 100.0 ^oC (The boiling point of water), then turn to steam, then heat from 100.0 ^oC to 175 ^oC.  Add all the heats up.

 

First - use

m = 3.29 kg
DT = 79 ^oC (starts at 21.0^oC ends at 100 ^oC )
C = 4186 J/kg^oC (for the liquid)
Q = 1,087,983.26 J

 

Next use

m = 3.29 kg
L = 22.6 x 10⁵ J/Kg (vaporization - it's boiling)
Q = 7,435,400 J
 

Next - use

m = 3.29 kg
DT = 75 ^oC (starts at 100^oC ends at 175 ^oC )
C = 2010 J/kg^oC (steam)
Q = 495,967.5 J

Finally - add them all up:
1,087,983.26 J + 7,435,400 J + 495,967.5 J = 9,019,350.76 J = 9.02 x 10⁶ J

 

5. 500. grams of a mystery liquid at 45 ^oC is mixed with 300. grams of water (C = 4186 J/Kg^oC) initially at 22 ^oC.  The final temperature of the mixture is 33 ^oC.  What is the specific heat of the mystery liquid? (Assuming no heat was lost to the surroundings)

(2,302.3 – 2.30 x 10³ J/kg^oC)

 

Calorimetry problems are
m₁c₁DT₁ = m₂c₂DT₂

Sometimes there is more than one term on a side, but the left side is for the hot things that lose heat, and the right side is the things that gain the heat:

 m₁ = .500 kg

 c₁ = ??

 DT₁ = 45 - 33 = 12 ^oC

 

 m₂ = 0.300 kg

 c₂ = 4186 J/kg^oC

 DT₂ = 33 - 22 = 11 ^oC