Calculating Mass Using Specific Heat Capacity and Heat Received

The question asks us to find the mass of a material given its specific heat capacity, the amount of heat it receives, and the resulting change in temperature. This involves the fundamental concept of specific heat capacity in thermodynamics.

Understanding Specific Heat Capacity

Specific heat capacity ($c$) is a physical property of a substance that quantifies the amount of heat energy required to raise the temperature of 1 kilogram of that substance by 1 degree Celsius (or 1 Kelvin).

The relationship between the heat energy ($Q$) absorbed or released by a substance, its mass ($m$), its specific heat capacity ($c$), and the change in temperature ($\Delta T$) is given by the formula:

$$Q = mc\Delta T$$

In this problem, we are given $Q$, $c$, and the initial and final temperatures (from which we can find $\Delta T$). We need to find $m$.

Given Information:

• Specific heat capacity ($c$) = 400 J/(kg °C)
• Initial temperature ($T_1$) = 15 °C
• Final temperature ($T_2$) = 25 °C
• Heat received ($Q$) = 20 kJ = 20,000 J (Note: 1 kJ = 1000 J)

Finding the Change in Temperature ($\Delta T$)

The change in temperature is the difference between the final temperature and the initial temperature.

$$\Delta T = T_2 - T_1$$

$$\Delta T = 25 \text{ °C} - 15 \text{ °C}$$

$$\Delta T = 10 \text{ °C}$$

Rearranging the Formula to Find Mass

We use the formula $Q = mc\Delta T$ and rearrange it to solve for mass ($m$).

$$m = \frac{Q}{c\Delta T}$$

Calculating the Mass

Now, we substitute the given values into the rearranged formula:

$$m = \frac{20000 \text{ J}}{400 \text{ J/(kg °C)} \times 10 \text{ °C}}$$

$$m = \frac{20000 \text{ J}}{4000 \text{ J/kg}}$$

$$m = \frac{20000}{4000} \text{ kg}$$

$$m = 5 \text{ kg}$$

Conclusion

The mass of the material is 5 kg.

Revision Table: Key Concepts in Specific Heat

Additional Information on Thermal Properties

• Different materials have different specific heat capacities. Water, for instance, has a very high specific heat capacity (about 4186 J/(kg °C)), which is why it is used in cooling systems and takes a long time to heat up or cool down.
• Materials with low specific heat capacity heat up and cool down quickly. Metals like copper and aluminum have lower specific heat capacities compared to water.
• The formula $Q = mc\Delta T$ assumes that the specific heat capacity remains constant over the given temperature range, which is generally true for small temperature changes.
• Heat transfer can occur via conduction, convection, and radiation. This problem deals with the effect of transferred heat energy on the internal energy (and thus temperature) of the material.