This preview shows page 1. Sign up to view the full content.
Unformatted text preview: bly put an ice cube into a hot drink to cool it
quickly. You were engages, in a somewhat trialanderror way, in a practical aspect of heat
transfer known as calorimetry.
Figure 2 shows two systems thermally
interacting with each other but isolated from
System 1
everything else. Suppose they start at different
temperatures T1 and T2. As you know from
Qnet = Q1 + Q2 = 0
T1
experience, eat energy will be transferred from the
hotter to the colder system until they reach a
common final temperature Tf. The systems will
Q1
then be in thermal equilibrium and the
Q2
System 2
temperature will not change further.
The insulation prevents any heat energy
T2
from being transferred to or from the
environment, so energy conservation tells us that
any energy leaving the hotter system must enter
the colder system. That is, the systems exchange
energy with no net less or gain. The concept is
straightforward, but to state the idea Figure 2 Two systems interacting thermally.
mathematically we need to be careful with signs.
Let Q1 be the energy transferred to system 1 as heat. Q1 is positive if energy enters
system 1, negative if energy leaves system 1. Similarly, Q2 is the energy transferred to system 2.
The fact that the systems are merely exchanging energy can be written as
    (11) That is, the energy lost by the hotter system is the energy gained by the colder system. But Q1
and Q2 have opposite signs, so
(12)
Because no energy is exchanged with the environment, it makes more sense to write this
relationship as
0 6 (13) This idea is not limited to the interaction of only two systems. If three or more systems are
combined in isolation from the rest of their environment, each at a different initial temperature, they will
all come to a common final temperature that can be found from the relationship 0 (14) NOTE: The signs are very important in calorimetry problems. ΔT is always
Tf Ti, so ΔT and Q are negative for any system whose temperature decreases.
The proper sign of Q for any phase change must be supplied by you, depending
on the direction of the phase change.
2.8 Problem Solving Strategy
Model
Visualize
Solve •
•
• Assess Identify the interacting systems. Assume that they are isolated from the larger environment.
List known information and identify what you need to find. Convert all quantities to SI
units.
The mathematical representation, which is a statement of energy conservation, is Qnet = Q1 + Q2 + Q3 + Q4 + L = 0 For systems that undergo a temperature change, Q = Mc(TfTi). Be sure to have the temperatures
in the correct order.
For systems that undergo a phase change, Q = ±ML. Supply the correct sign by observing
whether energy enters or leaves the system during the transition.
Some systems may undergo a temperature change and a phase change. Treat the changes
separately. The heat energy is Q = QΔT+Qphase.
Is the final temperature in the middle? Tf that is higher or lower than all temperatures is an
indication that something is wrong, usually a sign error. 3.0 References
•
•
• An Introduction to Thermal Physics by Daniel V. Schroeder. Addison Wesley Publishing
Fundamentals of Physics extended 7th edition by Halliday, Resnick, and Walker. Wiley
Publishing
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics by Randall D.
Knight. Addison Wesley Publishing. 4.0 The Experiment
The experiment that you will perform today is measuring the specific heat of a metal using a
calorimeter. We start with the fact the specific heat of a metal is given by (15) ∆ where m is the mass of the sample, Q is the heat of the sample, and ΔT is the change in the temperature 7 measured in Celsius or Kelvin.
In the experiment that you will perform today, there are three different systems: the water, shot,
and cup. The shot will be heated up by placing it in boiling water. Thus the shot will be gaining heat.
Using equation (4), we can write the equation for the heat loss as (16)
Once the shot is heated, you will place it inside a cup and cool water. Thus the heat will be transferred to
the cup and water and the shot will be losing heat. We know from equation (4) that the energy gained is (17)
Using the conservation of energy, equation (12), (18)
(19) (20) 8...
View
Full
Document
This document was uploaded on 03/20/2014 for the course PHYS 215 at Lafayette.
 Fall '09
 Heat

Click to edit the document details