Calculate the heat gained by the calorimeter and water, given their masses, initial temperatures, specific heat capacities, and the final mixture temperature.

The calorimeter starts at the same temperature as the water.

$$\Delta T_c = T_f - T_{w,i} = 34\,^{\circ}\text{C} - 25\,^{\circ}\text{C} = 9\,^{\circ}\text{C} = 9\, \text{K}$$

$$Q_c = C_c \Delta T_c$$ $$Q_c = (40\, \text{J/K}) (9\, \text{K})$$ $$Q_c = 360\, \text{J}$$

The heat gained by the calorimeter is $\boxed{\text{360 J}}$.

c) (ii) Heat gained by water;

$$Q_w = m_w c_w \Delta T_w$$ $$Q_w = (0.1\, \text{kg}) (4200\, \text{J/kg}\cdot\text{K}) (9\, \text{K})$$ $$Q_w = 3780\, \text{J}$$

The heat gained by water is $\boxed{\text{3780 J}}$.

c) (iii) Heat lost by the metal block;

Heat lost by the metal block equals the total heat gained by the water and the calorimeter.

$$Q_m = Q_w + Q_c$$ $$Q_m = 3780\, \text{J} + 360\, \text{J}$$ $$Q_m = 4140\, \text{J}$$

The heat lost by the metal block is $\boxed{\text{4140 J}}$.

c) (iv) Specific heat capacity of the metal block

$$Q_m = m_m c_m \Delta T_m$$ $$4140\, \text{J} = (0.15\, \text{kg}) c_m (66\, \text{K})$$ $$4140\, \text{J} = (9.9\, \text{kg}\cdot\text{K}) c_m$$

$$c_m = \frac{4140\, \text{J}}{9.9\, \text{kg}\cdot\text{K}}$$ $$c_m = 418.1818... \, \text{J/kg}\cdot\text{K}$$

Rounding to one decimal place:

$$c_m = 418.2\, \text{J/kg}\cdot\text{K}$$

The specific heat capacity of the metal block is $\boxed{\text{418.2 J/kg}\cdot\text{K}}$.

Question 14:

a) Distinguish between solid and liquid states of matter in terms of intermolecular forces.

In solids, intermolecular forces are very strong, holding particles in fixed positions and giving solids a definite shape and volume. In liquids, intermolecular forces are weaker than in solids, allowing particles to move past each other, giving liquids a definite volume but no definite shape.

b) In an experiment to estimate the diameter of an oil molecule, an oil drop of diameter $0.05\, \text{cm}$ spreads over a circular patch whose diameter is $20\, \text{cm}$.

b) (i) Determine the volume of the oil drop

$$V_{drop} = \frac{4}{3} \pi r_{drop}^3$$ $$V_{drop} = \frac{4}{3} \times \pi \times (0.025\, \text{cm})^3$$ $$V_{drop} = \frac{4}{3} \times \pi \times 0.000015625\, \text{cm}^3$$ $$V_{drop} \approx 6.54498 \times 10^{-5}\, \text{cm}^3$$

Rounding to three significant figures:

$$V_{drop} = 6.54 \times 10^{-5}\, \text{cm}^3$$

The volume of the oil drop is $\boxed{\text{6.54} \times 10^{-5}\, \text{cm}^3}$.

b) (ii) Determine the area of the patch covered by the oil

$$A_{patch} = \pi r_{patch}^2$$ $$A_{patch} = \pi \times (10\, \text{cm})^2$$ $$A_{patch} = \pi \times 100\, \text{cm}^2$$ $$A_{patch} \approx 314.159\, \text{cm}^2$$

Rounding to three significant figures:

$$A_{patch} = 314\, \text{cm}^2$$

The area of the patch covered by the oil is $\boxed{\text{314 cm}^2}$.

b) (iii) Determine the diameter of an oil molecule (implied by the context of the experiment)

The volume of the oil drop is equal to the volume of the cylindrical patch.

$$V_{drop} = A_{patch} \times D_{molecule}$$

$$D_{molecule} = \frac{V_{drop}}{A_{patch}}$$ $$D_{molecule} = \frac{6.54498 \times 10^{-5}\, \text{cm}^3}{314.159\, \text{cm}^2}$$ $$D_{molecule} \approx 2.0833 \times 10^{-7}\, \text{cm}$$

Rounding to three significant figures:

$$D_{molecule} = 2.08 \times 10^{-7}\, \text{cm}$$

The diameter of an oil molecule is $\boxed{\text{2.08} \times 10^{-7}\, \text{cm}}$.