Calculate the Thermal Capacity of a Fluid
What is the Thermal Capacity of a liquid, given the following details:
Volume = 200ml
Specific Gravity = 0.8
Specific Heat = 1.5 cal/g/degC
Also, assuming there are no external heat losses, calculate the energy required to raise the sample's temperature to 37 degC.
The Thermal Capacity of a sample is the amount of heat energy required to raise the specific sample by one degree celcius. This can calculate from the given data as follows:
Weight of Sample = Specific Gravity * Weight of Same Volume of Water
Weight of Sample = 0.8 * 200 g
Weight of Sample = 160g
Water has a density of one gram per milliliter.
Then:
Thermal Capacity = Weight of Sample * Specific Heat
Thermal Capacity = 1.5 cal/g/degC * 160g
Thermal Capacity = 240 cal/degC
It is reassuring to see that the final result has the correct units to be a thermal capacity - an amount of energy per unit of temperature!
The thermal capacity we just calculated is the energy required to raise this samples temperature by one degree celcius. Now, we want to go from 22 to 37 degC, so we need to multiply the thermal capacity by 15 degC to get the number of calories required.
Energy = Thermal Capacity * Temperature Change
Energy = 240 cal/degC * 15 degC
Energy = 3600 cal
Again, we are reassured by the fact that the units work out correctly.
In fact, it should be noted that this entire problem can be done with very little physics knowledge - pure application of dimensional analysis to the quantities given and wanted (and their units) can quickly solve this problem!
Determine the Efficiency of the Carnot Cycle
Determine the Efficiency of the Carnot Cycle assuming the piston is filled with one mole of a monatomic ideal gas.First of all, we are after the efficiency of a thermodynamic cycle, E, which is given by:
E=W/Qin
where W is the work done by the cycle and Qin is the heat that flows into the engine.
To proceed, we calculate the work done and the heat absorbed by the system on each of the four legs of the cycle:The gas is an ideal gas, so it obeys the ideal gas law: PV=nRT where P is presure, V is volume, n is the number of moles of gas, R is the gas constant and T is the temperature.
Then, the work done on a->b is given by:Wab=∫P dV
Wab=RTh∫ dV/V
Wab=RTh ln(Vb/Va)
Similarly, Wcd = RTl ln(Va/Vb) (this is negative, so on c->d the outside does work on the system)Wbc and Wda are both zero as there is no change int he volume of the system on these steps.
Also, as the gas is an ideal, monatomic gas, the specific heat at constant volume, Cv =3/2 R.
Then, the heat flow on the b->c and d->a legs is given by:
Q=nCvΔT
Qbc=3R(Tl-Th)/2
and
Qda=3R(Th-Tl)/2
Also, for an ideal gas the Internal Energy, U=3/2RT, so as the temperature on the isothermal legs does not change, exactly enough heat must flow in or out of the system to keep the energy constant while the work is being done.
Thus, the total work done, W, is given by:
W= R(Th-Tl)ln(Vb/Va)
Heat only flows in to the system (ie Q positive) on the a->b and d->a legs - thus, Qin is given by:
Qin=3/2 R(Th-Tl) + RTh ln(Vb/Va)
Then
E= (Th-Tl)Rln(Vb/Va) / (3/2 R(Th-Tl) + RTh ln(Vb/Va))
While the Carnot engine is always less efficient that the Carnot engine, the difference becomes small for Vb>>Va and Th>>Tl
Heat Capacity Calculations
1. Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 56oC.
q = m x Cg x (Tf - Ti)
m = 250g
Cg = 4.18 J oC-1 g-1
Tf = 56oC
Ti = 20oC
q = 250 x 4.18 x (56 - 20)
q = 250 x 4.18 x 36
q = 37 620 J = 38 kJ
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 Cg x (Tf - Ti)
q = 204.75 J
m = 15g
Ti = 25oC
Tf = 60oC
204.75 = 15 x Cg x (60 - 25)
204.75 = 15 x Cg x 35
204.75 = 525 x Cg
Cg = 204.75 ? 204.75= 0.39 JoC-1 g-1
3. 216 J of energy is required to raise the temperature of aluminium from 15o to 35oC. Calculate the mass of aluminium.
(Specific Heat Capacity of aluminium is 0.90 JoC-1g-1).
q = m x Cg x (Tf - Ti)
q = 216 J
Cg = 0.90 JoC-1g-1
Ti = 15oC
Tf = 35oC
216 = m x 0.90 x (35 - 15)
216 = m x 0.90 x 20
216 = m x 18
m = 216 ? 18 = 12g
Volume Calculations
1. What volume is needed to store 0.050 moles of helium gas at 202.6kPa and 400K?
PV = nRT
P = 202.6 kPa
n = 0.050 mol
T = 400K
V = ? L
R = 8.314 J K-1 mol-1
202.6V=0.050x8.314x400
202.6 V = 166.28
V = 166.28 ? 202.6
V = 0.821 L (821mL)
2. At a certain temperature, 1.0 L of a gas has a pressure of 760 torr. What will the volume be if the pressure changes to 380 torr?
a. What are the four variables that we use to measure a sample of gas?
volume, pressure, temperature, and number of moles
b. What is the relationship between volume and pressure?
If the temperature and number of moles remain constant, volume is inversely proportional to pressure.
c. Qualitatively, if the pressure of a gas increases (at constant temperature and number of moles), what happens to the volume?
the volume decreases
d. If the pressure doubles, how much will the volume decrease?
If the pressure goes up by a factor of 2, the volume will decrease by a factor of 2.
e. For our 1.0 L sample of gas, what will the new volume be?
760 torr / 380 torr x 1.0 L = 2.0 L
|
