# 1. Dermatologists often remove small precancerous skin lesions by freezing them... (a)

```PHYS1114 SAMPLE EXAM 5 SOLUTIONS
1.
Spring 2013
Professor Kenny L. Tapp
Dermatologists often remove small precancerous skin lesions by freezing them quickly with
liquid nitrogen, which has a temperature of 77 K. What is this temperature on the (a) Celsius
and (b) Fahrenheit scales?
a. The Kelvin temperature and the temperature on the Celsius scale are related by Equation 12.1: T
= Tc + 273.15, where T is the Kelvin temperature and
is the Celsius temperature. Therefore, a
temperature of 77 K on the Celsius scale is
b. The temperature of –196 °C is 196 Celsius degrees below the ice point of 0 °C. Since
this number of Celsius degrees corresponds to
,
Subtracting 353 Fahrenheit degrees from the ice point of 32.0 °F on the Fahrenheit scale gives a
Fahrenheit temperature of
.
2.
An ice chest at a beach party contains 12 cans of soda at 5.0 °C. Each can of soda has a
mass of 0.35 kg and a specific heat capacity of 3800 J/(kg?C°). Someone adds a 6.5-kg
watermelon at 27 °C to the chest. The specific heat capacity of watermelon is nearly the same
as that of water. Ignore the specific heat capacity of the chest and determine the final
temperature T of the soda and watermelon.
We assume that no heat is lost through the chest to the outside. Then, energy conservation dictates
that the heat gained by the soda is equal to the heat lost by the watermelon in reaching the final
temperature . Each quantity of heat is given by the equation,
, where we write the
change in temperature
as the higher temperature minus the lower temperature.
Starting with the statement of energy conservation, we have
Heat gained by soda = Heat lost by watermelon
Qsoda = Qwatermelon
(cmΔT) soda = (cmΔT) watermelon
€
Since the watermelon is being treated like water, we take the specific heat capacity of water. Thus,
the above equation becomes
€
Solving for
, we obtain
1
3.
b)
How much heat in kilocalories is required to change 2.3 kg of water from 38°C to 45°C?
4.
When you drink cold water, your body must expend metabolic energy in order to maintain
normal body temperature (37 °C) by warming up the water in your stomach. Could drinking ice
water, then, substitute for exercise as a way to “burn calories?” Suppose you expend 430
kilocalories during a brisk hour-long walk. How many liters of ice water (0 °C) would you have
to drink in order to use up 430 kilocalories of metabolic energy? For comparison, the stomach
can hold about 1 liter.
The volume V of a mass m of water is given by
, where r is the mass density of water. In
order to warm a mass m of ice water to body temperature, the body must provide an amount Q of
heat given by
, where c is the specific heat of water, and ΔT is 37 C°, the difference
between body temperature (37 °C) and the temperature of ice water (0 °C) . We will calculate the
required mass m of ice water, and then find the corresponding volume V.
The volume V of the water is
& solving
for the mass m, we obtain
The amount Q of heat is given as 430 kcal, which must be converted to joules via 1 kcal = 4186 J:
Solve for the volume of ice water in m3:
To convert this result to liters, we use the equivalence 1 liter =
:
2
5.
A 20 g iron nail is being pounded by a 0.44 kg hammer. The speed of the hammer as it strikes
the nail is 9.0 m/s. Assuming half the kinetic energy of the hammer is converted into thermal
energy of the nail, how many times should the nail be stuck to raise its temperature by 17°C?
(specific heat of iron = 448 J/kg °C).
3
6.
Find the total heat required to change 2.0 kg of ice at -10°C into water at 20°C. (Specific heat
for ice = 2090 J/kg °C, Latent Heat of Fusion = 3.34 x 105 J/kg, Latent Heat of Vaporization =
22.6 x 105 J/kg).
7.
Define and provide an example for each of the three methods of heat transfer.
CONDUCTION – CONVECTION - RADIATION
4
8.
Suppose that the steel gas tank in your car is completely filled when the temperature is 17 °C.
How many gallons will spill out of the twenty-gallon tank when the temperature rises to 35 °C?
Both the gasoline and the tank expand as the temperature increases. The coefficients of volumetric
expansion βg and βs for gasoline and steel can be found in any standard table of data. According to
the equation, the volume expansion of the gasoline is
while the volume of the steel tank expands by an amount
The amount of gasoline which spills out is
9.
One early May morning during finals week at the Oklahoma City Community College, the
temperature fell from 12.0°C to -20.0 °C in 27.0 minutes. Suppose that a 19-m aluminum
flagpole were subjected to this temperature change. Find the average speed at which its height
would decrease, assuming the flagpole responded instantaneously to the changing
temperature.
The average speed
of the flagpole’s contraction is given by
, where ΔL is the amount by
which it contracts, and Δt is the elapsed time. The amount of contraction the pole undergoes is found
from
, where a is the coefficient of thermal expansion for aluminum, L0 is the length of
the pole before it begins contracting, and ΔT is the difference between the higher and lower
temperatures of the pole.
Substituting
into
, we get…
The elapsed time Δt is given in minutes, which must be converted to seconds:
Substituting this result and the given values into our master equation, we obtain
END OF SAMPLE EXAM 5
5
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