[5 Mark Question Answer]

Question 15 Marks

Describe a method to determine the specific heat capacity of a solid, like a piece of copper.

Answer

The given solid is weighed and then heated by placing it in a beaker containing boiling water. The steady temperature of the solid is noted. A calorimeter with stirrer is weighed. The calorimeter is then filled with water and weighed again. Thus, the mass of water used is calculated. Initial temperature of water is noted. Solid is then transferred into calorimeter. The contents are stirred and final temperature is noted.
Mass of calorimeter with stirrer $=\mathrm{m}_1 \mathrm{~g}$
Specific heat capacity of calorimeter $=\mathrm{C}_1$ (given)
Mass of water taken $=\mathrm{m}_1 \mathrm{~g}$
Specific heat capacity of water $=c_{\text {, }}$ (given)
Mass of solid $=m_1 g$
Specific heat capacity of the solid (to be determined) $=c_J$
Initial temperature of the solid $=x^{\circ} \mathrm{C}$
Initial temperature of water + Calorimeter $=y^{\circ} \mathrm{C}$
Final temperature of the mixture $=z^{\circ} \mathrm{C}$
Heat lost by the solid $=$ Heat gained by the calorimeter and water
$ \begin{aligned} & m_3 C_3(x-z)=m 1 C_1(z-y)+m_2 C_2(z-y) \\ & =\left(m_1 C_1+m_2 C_2\right)(z-y) \\ & \mathrm{C}_3=\frac{\left(\mathrm{m}_1 \mathrm{C}_1+\mathrm{m}_2 \mathrm{C}_2\right)(\mathrm{z}-\mathrm{y})}{\mathrm{m}_3(\mathrm{x}-\mathrm{y})} \\ & \end{aligned} $
i.e. the specific heat capacity of the solid is calculated.

View full question & answer→

Question 25 Marks

A copper calorimeter weighing 57.5 g contains 60 g of water at 12°C. 55 g of iron nails at 100°C are dropped into the calorimeter ands stirred rapidly. The final temperature attained by the calorimeter and its contents is 20°C. Calculate the specific heat of iron. (Sp. heat of copper = 0.4 J/g°C, Sp. heat of water = 4.2 J/g°C).

Answer

Mass of calorimeter, $\mathrm{m}_1=57.5 \mathrm{~g}$
Specific heat capacity of calorimeter, $\mathrm{C}_1=0.4 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$
Mass of water taken, $\mathrm{m}_2=60 \mathrm{~g}$
Specific heat capacity of water, $C_z=4.2 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$
Mass of iron nails, $\mathrm{m}_2=55 \mathrm{~g}$
Specific heat capacity of iron $=\mathrm{C}_3$
Initial temperature of iron nails, $x=100^{\circ} \mathrm{C}$
Initial temperature of calorimeter + water, $y=12^{\circ} \mathrm{C}$
Final temperature of the mixture, $z=20^{\circ} \mathrm{C}$
Heat lost by iron nails $=$ heat gained by calorimeter and water
$ \begin{aligned} & \mathrm{m}_3 \mathrm{C}_3(\mathrm{x}-\mathrm{z})=\mathrm{m}_1 \mathrm{C}_1(\mathrm{z}-\mathrm{y})+\mathrm{m}_2 \mathrm{C}_2(\mathrm{z}-\mathrm{y}) \\ & \mathrm{C}_3=\frac{\left(\mathrm{m}_1 \mathrm{C}_1+\mathrm{m}_2 \mathrm{C}_2\right)(\mathrm{z}-\mathrm{y})}{\mathrm{m}_3(\mathrm{x}-\mathrm{z})} \\ & =\frac{(57.5 \times 0.4+60 \times 4.2)(20-12)}{55 \times(100-20)}=0.5 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \end{aligned} $

View full question & answer→

Physics [5 Mark Question Answer]

Test yourself on this topic