Conceptual Physics

Variables

What is the strategy?

1. Use conservation of energy to state that the heat lost by the object equals the heat gained by the water bath. Apply the specific heat
   equation to write the conservation of energy equation in terms of the masses, temperatures, and specific heats.


2. Solve for the unknown specific heat, substitute the known values, and evaluate.

Physics principles and equations

By the conservation of energy, the heat gained by the water bath (beaker plus water) equals the heat lost by the object. The sum of the heat
transfers is zero.

Qw + Qo = 0

We use the equation below to relate the heat flow to specific heat.

Q = cmǻT

Step-by-step solution

We start with an equation stating the conservation of energy for this experiment. Then, the heat values in the equation are written with
expressions involving specific heat.

Now we solve the equation for the unknown specific heat of the object and evaluate.

Based on the values in the table of specific heats, it appears that the material may consist of aluminum.

water object

mass mw = 0.744 kg mo = 0.197 kg

initial temperature Tw = 23.2°C To = 67.8°C

specific heat cw = 4178 J/kg·K co

heat transferred to water Qw

heat transferred from object Qo

final temperature Tf = 25.6°C

Step Reason

1. Qw + Qo = 0

Qw = íQo

conservation of energy

2. Qo = comoǻT specific heat equation

3. Qo = como(Tf – To) initial and final temperatures for object

4. Qw = cwmw(Tf – Tw) specific heat equation

5. cwmw(Tf – Tw) = – como(Tf – To) substitute equations 3 and 4 into equation 1

Step Reason

6. solve for specific heat of object

7. substitute

8. co = 897 J/kg·K evaluate

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