SPECIFIC HEAT AND VISCOSITY OF
FLUID
• Specific Heat:
• Specific heat is the amount of heat energy
required to raise the temperature of one unit of
mass of a substance by one degree Celsius (or
Kelvin).
• It is typically measured in joules per kilogram per
degree Celsius (J/kg°C).
• The specific heat of a substance depends on its
composition and phase. For example, water has a
specific heat of 4.18 J/kg°C, while the specific heat
of air is much lower at around 1.00 J/kg°C.
Specific Heat
a. Some things heat up or cool down faster than others.
Land heats up and cools down faster than water
Specific heat is the amount of heat required to raise the temperature of
1 kg of a material by one degree (C or K).
1) C water = 4184 J / kg C
2) C sand = 664 J / kg C
This is why land heats up quickly during the day and cools quickly at
night and why water takes longer.
Why does water have such a high specific heat?
water
metal
Water molecules form strong bonds with each other; therefore it
takes more heat energy to break them. Metals have weak bonds
and do not need as much energy to break them.
How to calculate the amount of heat energy
required to raise the tmperature of a
substance?
Q = m * c * ΔT
Q =Heat energy, measured in joules (J)
m = Mass, measured in kilograms (kg)
T =Change in temperature, measured in Celsius
(°C)
C = Specific heat, measured in joules per
kilogram Celsius (J/kg°C)
Some common uses of specific heat in daily life
include:
• Cooking: Specific heat is used in cooking to determine how
much heat energy is required to cook food to a desired
temperature.
• Heating and cooling systems: Specific heat is used in
designing heating and cooling systems for buildings and
other structures.
• Industrial processes: Specific heat is used in various
industrial processes, such as heating and cooling of
chemicals and materials.
If the specific heat capacity of ice is 0.5
cal/gC°, how much heat would have to be
added to 200 g of ice, initially at a
temperature of -10°C, to raise the ice to
the melting point?
a)
b)
c)
d)
1,000 cal
2,000 cal
4,000 cal
0 cal
m = 200 g
c = 0.5 cal/gC°
T = -10°C
Q = mcT
= (200 g)(0.5 cal/gC°)(10°C)
= 1,000 cal
(heat required to raise the temperature)
example: If the specific heat capacity of
ice is 0.5 cal/gC°, how much heat would
have to be added to 200 g of ice, initially at
a temperature of -10°C, to completely melt
the ice? (Latent heat is 80 cal/g)
Lf = 80 cal/g
a)
b)
c)
d)
1,000 cal
14,000 cal
16,000 cal
17,000 cal
Q = mLf
= (200 g)(80 cal/g)
= 16,000 cal
(heat required to melt the ice)
Total heat required to raise the ice to 0 °C and then to melt
the ice is:
1,000 cal + 16,000 cal = 17,000 cal = 17 kcal
A hot plate is used to transfer 400 cal
of heat to a beaker containing ice and
water; 500 J of work are also done on
the contents of the beaker by stirring.
What is the increase in internal energy
of the ice-water mixture? (note: 1 cal =
4.19J)
a)
b)
c)
d)
900 J
1180 J
1680 J
2180 J
W = -500 J
Q = 400 cal
= (400 cal)(4.19 J/cal)
= 1680 J
U = Q - W
= 1680 J - (-500 J)
= 2180 J
Viscosity
• is the quantity that describes a fluid's
resistance to flow.
• Viscosity is a measure of a fluid's resistance to
flow. It is typically measured in units of poise
or centipoise (1 poise = 0.1 Pa·s). The viscosity
of a fluid depends on its composition,
temperature, and pressure.
• For example, honey has a much higher
viscosity than water, which means it flows
much more slowly.
Problem: A fluid has a viscosity of 0.01 Pa·s and
flows through a pipe with a diameter of 10 cm at a
velocity of 2 m/s. What is the flow rate of the fluid?
Solution:
The flow rate (Q) of the fluid can be calculated using the formula:
Q=A*v
where A is the cross-sectional area of the pipe and v is the velocity of the fluid.
The cross-sectional area of the pipe can be calculated using the formula for the
area of a circle:
A = π * (d/2)^2
where d is the diameter of the pipe.
Substituting the given values, we get:
A = π * (10 cm/2)^2 = 78.54 cm^2
v = 2 m/s
Q = A * v = 78.54 cm^2 * 2 m/s = 157.08 cm^3/s
Therefore, the flow rate of the fluid is 157.08 cm^3/s.
Problem: A fluid has a viscosity of 0.02 Pa·s and flows
through a pipe with a length of 10 m and a diameter of 5
cm. If the pressure drop across the pipe is 5 kPa, what is
the flow rate of the fluid?
Solution:
The flow rate (Q) of the fluid can be calculated using the Poiseuille's Law formula:
Q = (π * ΔP * r^4) / (8 * η * L)
where ΔP is the pressure drop across the pipe, r is the radius of the pipe, η is the viscosity of
the fluid, and L is the length of the pipe.
The radius of the pipe can be calculated using the formula:
r = d/2
where d is the diameter of the pipe.
Substituting the given values, we get:
d = 5 cm = 0.05 m
r = 0.025 m
ΔP = 5 kPa = 5000 Pa
η = 0.02 Pa·s
L = 10 m
Q = (π * ΔP * r^4) / (8 * η * L) = (π * 5000 Pa * (0.025 m)^4) / (8 * 0.02 Pa·s * 10 m) = 0.049
m^3/s
Therefore, the flow rate of the fluid is 0.049 m^3/s.