Answer:
[tex]12.3 ^\circ C[/tex]
Explanation:
Conservation of Energy
The law states that energy cannot be created nor destroyed, only transferred. In an equation, it is
[tex]E_{initial} = E_{final}[/tex].
[tex]\dotfill[/tex]
Heat
Heat is the transfer of thermal energy between two materials that is due to the differences of temperatures. Between the two objects, the one with a higher temperature transfers heat onto the colder object.
An equation can be used to calculate the amount of heat gained or lost.
[tex]Q=mc\Delta T[/tex],
where
- Q is heat (joules)
- m is the mass of the material (grams)
- c is the specific heat capacity (J/(g × C°))
- T is the change in temperature (Celsius)
[tex]\hrulefill[/tex]
Solving the Problem
We're told that two objects are in contact with each other.
- 0.225 kg of iron
- 0.15 kg of water
- -28.7°C for iron
- 18.9°C for water
- iron's c value is 452
- water's c value is 4186
Water is the hotter object, so the thermal energy that it expels is the thermal energy that is gained by the block of iron.
Thus,
[tex]E_{water,lost}= E_{iron,gain}[/tex].
We can substitute both E values with the heat equation since the energies are being transferred.
[tex]m_{water}c_{water}\Delta T=m_{iron}c_{iron} \Delta T[/tex]
[tex]\dotfill\\[/tex]
Temperature
The problem states that after contact the iron and water reach an equilibrium temperature, which is what we're finding, so their final temperatures are the same.
Since the water is cooling down, its temperature change can be rewritten as the difference between the final temperature and 18.9°C.
For iron, since it's heating up, the temperature change is the difference between 28.7°C and the final temperature.
[tex](0.15)(4186)(18.9-T_f)=(0.225)(452)( T_f-(-28.7))[/tex]
(Although mass is standardized to be in grams because the specific heat capacity is in kg, we can leave mass as it is.)
Now, we solve for the final temperature.
[tex](0.15)(4186)(18.9-T_f)=(0.225)(452)( T_f+28.7)[/tex]
[tex](627.9)(18.9-T_f)=(101.7)( T_f+28.7)[/tex]
[tex]11867.31-627.9T_f=101.7T_f+2918.79[/tex]
[tex]8948.52=729.6T_f[/tex]
[tex]12.3^\circ C \approx T_f[/tex]
