Class 9 - Unit # 9 : Thermal Properties of Matter - Solved Numericals

Temperature Scales

1. c) Convert $30^\circ\text{C}$ into Kelvin and Fahrenheit Scale.

Data:

• Temperature in Celsius ($T_C$) = $30^\circ\text{C}$
• Temperature in Fahrenheit ($T_F$) = ?
• Temperature in Kelvin ($T_K$) = ?

Solution:

Part 1: Conversion to Kelvin

$$T_K = T_C + 273.15$$ $$T_K = 30 + 273.15 = 303.15\text{ K}$$

Part 2: Conversion to Fahrenheit

$$T_F = \left(\frac{9}{5} \times T_C\right) + 32$$ $$T_F = \left(\frac{9}{5} \times 30\right) + 32$$ $$T_F = 54 + 32 = 86^\circ\text{F}$$

RESULT: The temperature is $303.15\text{ K}$ on the Kelvin scale and $86^\circ\text{F}$ on the Fahrenheit scale.

2. c) Convert $212^\circ\text{F}$ into Celsius and Kelvin.

Data:

• Temperature in Fahrenheit ($T_F$) = $212^\circ\text{F}$
• Temperature in Celsius ($T_C$) = ?
• Temperature in Kelvin ($T_K$) = ?

Solution:

Part 1: Conversion to Celsius

$$T_C = \frac{5}{9} \times (T_F - 32)$$ $$T_C = \frac{5}{9} \times (212 - 32)$$ $$T_C = \frac{5}{9} \times 180 = 5 \times 20 = 100^\circ\text{C}$$

Part 2: Conversion to Kelvin

$$T_K = T_C + 273.15$$ $$T_K = 100 + 273.15 = 373.15\text{ K}$$

RESULT: The temperature is $100^\circ\text{C}$ on the Celsius scale and $373.15\text{ K}$ on the Kelvin scale.

Specific Heat Capacity

3. c) How much heat is required to boil $3\text{ kg}$ of water which is initially at $10^\circ\text{C}$?

Data:

• Mass of water ($m$) = $3\text{ kg}$
• Initial Temperature ($T_1$) = $10^\circ\text{C}$
• Final Temperature ($T_2$) = $100^\circ\text{C}$ (boiling point of water)
• Change in Temperature ($\Delta T$) = $T_2 - T_1 = 100 - 10 = 90^\circ\text{C}$ (or $90\text{ K}$)
• Specific heat capacity of water ($c$) = $4200\text{ J/(kg}\cdot\text{K)}$
• Heat required ($\Delta Q$) = ?

Solution:

According to the equation for heat transfer:

$$\Delta Q = m \cdot c \cdot \Delta T$$ $$\Delta Q = 3\text{ kg} \times 4200\text{ J/(kg}\cdot\text{K)} \times 90\text{ K}$$ $$\Delta Q = 12,600 \times 90 = 1,134,000\text{ J}$$

RESULT: The heat required to boil the water is $1,134,000\text{ J}$ (or $1.134\text{ MJ}$).

4. b) $2\text{ kg}$ of copper requires $2050\text{ J}$ of heat to raise its temperature through $10^\circ\text{C}$. Calculate the heat capacity and specific heat capacity of the sample.

Data:

• Mass of copper ($m$) = $2\text{ kg}$
• Heat supplied ($\Delta Q$) = $2050\text{ J}$
• Change in Temperature ($\Delta T$) = $10^\circ\text{C}$
• Heat Capacity ($C$) = ?
• Specific Heat Capacity ($c$) = ?

Solution:

Part 1: Heat Capacity ($C$)

$$C = \frac{\Delta Q}{\Delta T} = \frac{2050\text{ J}}{10\text{ K}} = 205\text{ J/K}$$

Part 2: Specific Heat Capacity ($c$)

$$c = \frac{\Delta Q}{m \cdot \Delta T} = \frac{2050\text{ J}}{2\text{ kg} \times 10\text{ K}}$$ $$c = \frac{2050}{20} = 102.5\text{ J/(kg}\cdot\text{K)}$$

RESULT: The heat capacity of the sample is $205\text{ J/K}$ and its specific heat capacity is $102.5\text{ J/(kg}\cdot\text{K)}$.

Thermal Expansion

13. An iron block of volume $3\text{ m}^3$ is heated, so that its temperature changes from $25^\circ\text{C}$ to $100^\circ\text{C}$. If the coefficient of linear expansion of iron is $11 \times 10^{-6}\;^\circ\text{C}^{-1}$, what will be the new volume of the iron block after heating?

Data:

• Initial Volume ($V_1$) = $3\text{ m}^3$
• Initial Temperature ($T_1$) = $25^\circ\text{C}$
• Final Temperature ($T_2$) = $100^\circ\text{C}$
• Temperature Difference ($\Delta T$) = $100 - 25 = 75^\circ\text{C}$
• Coefficient of Linear Expansion ($\alpha$) = $11 \times 10^{-6}\;^\circ\text{C}^{-1}$
• Coefficient of Volumetric Expansion ($\beta$) = $3\alpha = 3 \times 11 \times 10^{-6} = 33 \times 10^{-6}\;^\circ\text{C}^{-1}$
• Final Volume ($V_2$) = ?

Solution:

According to the formula for volumetric thermal expansion:

$$V_2 = V_1(1 + \beta \Delta T)$$

Substituting the given values into the equation:

$$V_2 = 3 \times [1 + (33 \times 10^{-6} \times 75)]$$ $$V_2 = 3 \times [1 + 0.002475]$$ $$V_2 = 3 \times 1.002475$$ $$V_2 = 3.007425\text{ m}^3$$

RESULT: The new volume of the iron block after heating will be $3.007425\text{ m}^3$.