Specific Heat, Latent Heat and Calorimetry


Specific Heat
-------------

Q = mCΔT

Conservation of energy: Q_lost = Q_gained

m₁C₁(T - T₁) + m₂C₂(T - T₂) + m₃C₃(T - T₃) ... = 0

Ex 1.

There are 100 grams of water in a calorimeter (foam cup). The temp. of the water 
is 21.3 degrees C.  A piece of metal (69 grams) is heated to 99 degrees C in a 
beaker full of water then placed in the cup... the water temperature changed 
to 26.9 degrees C. What is the specific heat of the metal?

Q lost by the metal + Q gained by the water = 0

m₁C₁(T - T₁) + m₂C₂(T - T₂) = 0

0.069 * C_metal * (99 - 26.9) = .1 * 4184 * (26.9 - 21.3)

C_metal = 471 J/kg deg C

Ex 2.

A ceramic coffee cup, with m = 116g and c = 1090J/kg deg C, is initially at room 
temperature (24.0 degC). If 225g of 80.3 deg C coffee and 12.2g of 5.00 deg C cream 
are added to the cup, what is the equilibrium temperature of the system? 
Assume that no heat is exchanged with the surroundings, and that the specific 
heat of coffee and cream are the same as the specific heat of water (4180 J/Kg
deg C).

Q gained by cup + Q gained by the cream + Q lost by coffee = 0

m₁C₁(T - T₁) + m₂C₂(T - T₂) + m₃C₃(T - T₃) = 0

m₁C₁(T - T₁) + m₂C₂(T - T₂) = -m₃C₃(T - T₃)

                           = m₃C₃(T₃ - T)

0.116 * 1090 * (T - 24) + 0.0122 * 4180 * (T - 5) = 4180 * 0.225 * (80.3 - T)

=> T = 70.5 degC

Latent Heat
-----------



Q = mL_{FUSION OR VAPORIZATION}

Example 1.  
How much heat energy is required to change 500g of ice to water at 40 C? 

Specific heat capacity of water = 4120 J/kg.K
Latent heat of fusion of water is 335000 J/ kg 


Q_Total = mL_f + mcΔT

      = 0.5*335000 + 0.5*4120(40 - 0) = 169560 J

Example 2. 

In an experiment to measure the specific latent heat of fusion of ice, warm water 
was placed in an aluminium calorimeter. Crushed dried ice was added to the water. 
The following results were obtained:

c_cal = 790 J/kgK
c_water = 4180 Jkg/K
Mass of calorimeter = 77.2 g
Mass of water = 92.5 g
Initial temperature of water = 29.4 C
Temperature of ice = 0 C 
Mass of ice = 19.2 g
Final temperature of water = 13.2 C 

What is L_f?

Heat Gained = Heat Lost

m_iceL_f + m_icec_waterΔT_ice = m_calc_calΔT_water + m_waterc_waterΔT_water
 
0.192*L_f + 0.192*4180*(13.2 - 0) = 0.772*790*(29.4 - 13.2) + 0.925*4180*(29.4 - 13.2)

∴ L_f = 3.2 x 10⁵ J/kg

Example 3.

A copper calorimeter of mass 180g contains 450g of water and 50g of ice, 
all at 0 C. Dry steam is passed into the calorimeter until a certain temperature, 
T, is reached. The mass of the calorimeter and its contents at the end of the 
experiment increased by 25g.  If no heat was lost to the surroundings, find 
the final temperature, T.  Take the specific heat capacities of water and copper 
to be 4200 J/kg K and 390 J/kg K, respectively.  Take the latent heat of fusion 
of ice to be 3.36 x 10⁵ J/kg and the  latent heat of vaporisation 
of water to be 2.26 x 10⁶ J/kg. 

Energy released when steam condenses at 100 C is
Q_a = mL_f = 0.025*2.26 * 10⁶ = 5.65*10⁴ J   

Energy released when condensed steam cools from 100 C to temperature q is
Q_b  = mcΔT = 0.025*4200*(100 - T) = 1.05*10⁴ - 105T  

Energy absorbed by 50g of ice on melting at 0 C is
Q_c = mL_f = 0.05*3.36*10⁵ = 1.68*10⁴ J     

Energy absorbed by 500g of water (the original 450g of water and 50g of 
melted ice) in warming from 0 C to T  is
Q_d = 0.5*4200*T = 2100 T     

Energy absorbed by 180g of copper (the calorimeter) in warming from 0 C to T C is
Q_e = mcΔT = 0.18*390*T = 70.2T   

heat lost = heat gained

Q_a + Q_b = Q_c + Q_d + Q_e
5.65*10⁴ + 1.05*10⁴ - 105T = 1.68*10⁴ + 2100T + 70.2T
5.02*10⁴ = 2.275*10³T
∴ T  = 22 C

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