Question

A certain amount of heat is added to a mass of aluminum and its temperature is raised to 57 degrees celsius. Suppose the same amount of heat is added to the same of copper. How much does the temperature of the copper rise?

Answer 1

Answer on Question 57589, Physics – Molecular Physics | Thermodynamics

Question:

A certain amount of heat is added to a mass of aluminum and its temperature is raised to 57 degrees Celsius. Suppose the same amount of heat is added to the same mass of copper. How much does the temperature of copper rise?

Solution:

We can find the amount of heat that is added to a mass of aluminum from the formula:

$$Q _ {1} = m _ {1} c _ {1} \Delta t _ {1},$$

here, $m_{1}$ is the mass of aluminum, $c_{1} = 910\frac{J}{kg{}^{\circ}C}$ is the specific heat capacity of aluminum, $\Delta t_{1}$ is the change in the temperature.

Similarly, we can find the amount of heat that is added to the mass of the copper:

$$Q _ {2} = m _ {2} c _ {2} \Delta t _ {2},$$

here, $m_{2}$ is the mass of copper, $c_{2} = 390\frac{J}{kg{}^{\circ}C}$ is the specific heat capacity of copper, $\Delta t_{2}$ is the change in the temperature.

Since the amounts of heat $Q_{1}$ and $Q_{2}$ are the same, we can equate both expressions:

$$m _ {1} c _ {1} \Delta t _ {1} = m _ {2} c _ {2} \Delta t _ {2}.$$

Finally, we can find $\Delta t_{2}$ from the previous formula (since $m_{1} = m_{2}$ , the masses are canceled):

$$\Delta t _ {2} = \frac {c _ {1} \Delta t _ {1}}{c _ {2}} = \frac {9 1 0 \frac {J}{k g {}^ {\circ} C} \cdot 5 7 {}^ {\circ} C}{3 9 0 \frac {J}{k g {}^ {\circ} C}} = 1 3 3 {}^ {\circ} C.$$

Answer:

$$\Delta t _ {2} = 1 3 3 {}^ {\circ} \mathrm {C}.$$

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