TRIDENT UNIVERSITY
Module 5

Case
Heat and Thermodynamics
RAY WILLIAMS
4/24/2013
Ray Williams
April 20, 2013
Module 5

Case
Heat and Thermodynamics
1.
How much energy (in kcal )
is required to raise the temperature of 1 L of water from 20 to 100C,
and then turn it to steam at 100C?
VARIABLES:
q(joules)=1000grams
1 liter of water is 1000 milliliters
1L would weigh 1000 grams( ~ 36 oz )
q(joules) = mass * specific heat * (Temp.
final

Temp. initial)
q = 1000 grams * 4.180 J/gC * (100 C

20 C)
q = 334400 joules
And 1 cal = 4.186 J
So q in kcal, 79885.33 cal = 79.885 kcal
ANSWER: 79.885 kcal
2.
A 2 kg rod of aluminum (c = 0.9 kJ/kg

C) at 90C is dropped into 10 L of water at 10C
.
What is the final temperature of the mixture?
Heat lost by the rod = heat gained by the water
q_lost = q_gain
mass_aluminium * sp.heat_aluminium *(Tf
–
Ti) = mass_water * sp.heat_water*(Tf

Ti)
1L = 1000 milliliter
1L = 1000g, so 10L = 10000g
2kg*
0.9kJ/k
gC * (90C
–
x) = 10000g * 4.180 J/gC * (x
–
10C)
2000g * 0.9 J/gC * (90
–
x) = 10000g * 4.180 J/gC * (x
–
10C)
1800 (90
–
x) = 41800 ( x
–
10)
90
–
x = 23.222 (x
–
10)
90
–
x = 23.222x
–
232.22
23.222x + x = 232.222 + 90
24.222x = 322.222
X = 13.308 C
ANSWER
:
13.308C
Ray Williams
April 20, 2013
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 35C. What is the specific heat of the metal, in kcal/kg

C ?
Heat lost by the rod = heat gained by the water
q_lost =
q_gain
mass_metal * sp.heat_metal *(Tf
–
Ti) = mass_water * sp.heat_water*(Tf

Ti)
1L = 1000 milliliter
1L = 1000g,
0.2kg* x * (50C
–
35C) = 1000g * 4.180 J/gC * (35C
–
4C)
200g * x* (15C) = 1000g * 4.180 J/gC * (31C)
3000 gC * x
=
129580 J
x =
129580 J
/ 3000 gC
X
= 4
3.
19
3
3J/g
C
ANSWER:
43.1933 kJ/kgC
4.
A liter of gas, initially at a pressure of 500Pa, is compressed from 1.00 L to 0.25 L. During
the compression process, heat is dissipated to maintain a constant temperature. What is the final
pressure?
Pi = 500 Pa
Vi = 1L
Vf = 0.25L
PV = nRT
PiVi = nRT = 500Pa * 1L
Pf*Vf = nRT = Pf * 0.25L
Equating both,
500Pa * 1L = Pf * 0.25L
Pf = 2000Pa
Final pressure, Pf = 2000 Pa = 2 kPa
ANSWER: 2 kPa
5.
A radioisotope thermoelectric generator (RTG) generates elec
tricity by absorbing heat from a
core containing a radioactive material (plutonium 238 is typical) and discharging it into the
environment. An RTG installed on an unmanned sonobuoy, tethered to the ocean floor, has a
core temperature of 150 C, and discharg
es heat into sea water at 10 C. What is its theoretical
efficiency?
T_cold = 10C
T_hot = 150C
Theoretical efficiency = 1
–
Tcold / Thot
Theoretical effieciency = 1
–
(10/150) = 1

0.0667 = 93.333%
ANSWER:
93.333%
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