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# Applied physics i case study sample

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275 words
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\$ 0.00 ## Heat and Thermodynamics

Work out the following problems. Be sure to show your work in detail. See the ” Module 5 Handout” for examples of how the problems should be solved and presented.

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1. How much energy (in J) is required to raise the temperature of 500 g of water from 20 to 100C, and then turn it to steam at 100C?
Variables:
m = 500g = 0. 5 Kg
Increase in water temperature, ∆T = 100-20= 800C
Latent heat at 1000C, Lv = 2260 KJ/Kg
Specific heat capacity of water, c = 4. 185 kJ/Kg/0C
∆Q = ?
Relationships:
∆Q = m(c∆T + Lv)
Computation:
∆Q = 0. 5Kg x (4. 185 kJ/Kg/0C x 800C + 2260 KJ/Kg) = 1297. 4 kJ = 1297400J

2. A 2 kg rod of aluminum (c = 0. 9 kJ/kg-C) at 90C is dropped into 5 L of water at 4C. What is the final temperature of the mixture?
Variables:
mass of aluminium rod, mA = 2Kg
Volume of water, V = 5L = 0. 005 m3
Density of water, ρ = 1000 kg/m3
Mass of water, mw = ?
Specific heat capacity of aluminium, cA = 0. 9 kJ/Kg-C
Specific heat capacity of water, cw = 4. 185 kJ/Kg-C
initial temperature of rod, TiA = 90 0C
initial temperature of water, Tiw = 4 0C
Final equilibrium temperature of mixture, Tf = ?
Relationships:
mw = ρV
Assuming an adiabatic system, heat given by aluminium = heat taken by water, or,

mAcA(TiA-Tf) = mwcw(Tf-Tiw), or, Tf = (mAcATiA + mwcwTiw)/(mAcA + mwcw)
Computation:
mw = 1000 kg/m3 x 0. 005m3 = 5Kg

Tf = (2Kgx0. 9 kJ/kg-Cx900C + 5kgx4. 185kJ/kg-Cx40C)/( 2Kgx0. 9 kJ/kg-C +5kgx4. 185kJ/kgC)
= 10. 8 0C

3. A 0. 2 kg block of an unknown metal at 50C is immersed in 1L of water at 4C. The equilibrium temperature of the mixture is 25C. What is the specific heat of the metal, in kcal/kg-C ?
Varibles:
mass of metal, mm = . 2Kg
Volume of water, V = 1L = 0. 001 m3
Density of water, ρ = 1000 kg/m3
Mass of water, mw = ?
Specific heat capacity of water, cw = 1 kcal/kg-C
initial temperature of metal, Tim = 50 0C
initial temperature of water, Tiw = 4 0C
Final equilibrium temperature of mixture, Tf = 25 0C
Specific heat capacity of metal, cm = ?
Relationships:
mw = ρV
Assuming an adiabatic system, heat given by metal = heat taken by water, or,

mmcm(Tim-Tf) = mwcw(Tf-Tiw), or, cm = mwcw(Tf-Tiw)/(mm(Tim-Tf))
Computation:
mw = 1000 kg/m3 x 0. 001m3 = 1Kg
cm = 1kgx1kcal/kg-Cx(250C – 40C )/(0. 2kgx(500C-250C)= 4. 2 kcal/kg-C 3 hours! In only 3 hours, we’ll deliver you a flawless & plagiarism-free paper on any subject

4. A liter of gas, initially at a pressure of 500Pa, is compressed from 1. 00 L to 0. 30 L. During the compression process, heat is dissipated to maintain a constant temperature. What is the final pressure?
Variables:
Initial volume of gas, V1 = 1. 00 L
Final volume of gas, V2 = 0. 30 L
Initial pressure, P1 = 500 Pa
Final pressure, P2 = ?
For Isothermal process, T1 = T2
Relationships:
For isothermal process, P2 = P1/(V2/V1)
Computation:
P2 = 500 Pa/(0. 3L/1. 00L) = 1666. 67 Pa

5. A large thermoelectric array generates electricity by extracting heat from water at the surface of the ocean (25C) and dissipating it into the colder water at the bottom of the ocean (-2C). What is the maximum ideal efficiency of this arrangement?
Variables:
TC = -20C = 271K

TH = 25 0C = 298K
Maximum ideal efficiency, eff = ?

## Relationships: For a heat engine, maximum ideal efficiency,

Computation:
eff = 1-271K/298K = 0. 09 = 9%

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