Wolfram Alpha:
Search by keyword:
Astronomy
Chemistry
Classical Physics
Climate Change
Cosmology
Finance and Accounting
Game Theory
General Relativity
Lagrangian and Hamiltonian Mechanics
Macroeconomics
Mathematics
Microeconomics
Particle Physics
Probability and Statistics
Programming and Computer Science
Quantum Computing
Quantum Field Theory
Quantum Mechanics
Semiconductor Reliability
Solid State Electronics
Special Relativity
Statistical Mechanics
String Theory
Superconductivity
Supersymmetry (SUSY) and Grand Unified Theory (GUT)
The Standard Model
Topology
Units, Constants and Useful Formulas
Specific Heat
-------------
Q = mCΔT
Conservation of energy: Q_{lost} = Q_{gained}
m_{1}C_{1}(T - T_{1}) + m_{2}C_{2}(T - T_{2}) + m_{3}C_{3}(T - T_{3}) ... = 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
m_{1}C_{1}(T - T_{1}) + m_{2}C_{2}(T - T_{2}) = 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
m_{1}C_{1}(T - T_{1}) + m_{2}C_{2}(T - T_{2}) + m_{3}C_{3}(T - T_{3}) = 0
m_{1}C_{1}(T - T_{1}) + m_{2}C_{2}(T - T_{2}) = -m_{3}C_{3}(T - T_{3})
= m_{3}C_{3}(T_{3} - 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_{ice}L_{f} + m_{ice}c_{water}ΔT_{ice} = m_{cal}c_{cal}ΔT_{water} + m_{water}c_{water}ΔT_{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 10^{5} 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 10^{5} J/kg and the latent heat of vaporisation
of water to be 2.26 x 10^{6} J/kg.
Energy released when steam condenses at 100 C is
Q_{a} = mL_{f} = 0.025*2.26 * 10^{6} = 5.65*10^{4} J
Energy released when condensed steam cools from 100 C to temperature q is
Q_{b} = mcΔT = 0.025*4200*(100 - T) = 1.05*10^{4} - 105T
Energy absorbed by 50g of ice on melting at 0 C is
Q_{c} = mL_{f} = 0.05*3.36*10^{5} = 1.68*10^{4} 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*10^{4} + 1.05*10^{4} - 105T = 1.68*10^{4} + 2100T + 70.2T
5.02*10^{4} = 2.275*10^{3}T
∴ T = 22 C