A copper can of mass 0.2 kg contains 0.2kg of water at 10C. The can and its contents are placed in a refrigerator. Calculate the quantity of heat given out if the temperature of the can and its contents falls to -2C.

Here are the solutions to the questions:

The graph in the figure below shows a cooling curve of a liquid. Describe the main features of the curve. /3 marks

Step 1: Identify the initial phase.
The first sloping section shows the liquid cooling down, meaning its temperature is decreasing over time.

Step 2: Identify the phase change.
The flat horizontal section indicates a phase change (freezing). During this period, the liquid is turning into a solid, and the temperature remains constant at the freezing point, even though heat is being removed.

Step 3: Identify the final phase.
The second sloping section shows the substance in its solid state cooling down further, meaning its temperature is decreasing again.

C.I. Define the term specific latent heat of fusion. /1 mark
Specific latent heat of fusion is the amount of heat energy required to change a unit mass (typically 1 kg) of a substance from a solid state to a liquid state, or vice versa, at its melting/freezing point without any change in temperature.

II. A copper can of mass 0.2 kg contains 0.2 kg of water at 10°C. The can and its contents are placed in a refrigerator. Calculate the quantity of heat given out if the temperature of the can and its contents falls to -2°C. /3 marks
Given:
Mass of copper can ($m_{can}$) = $0.2 \text{ kg}$
Mass of water ($m_{water}$) = $0.2 \text{ kg}$
Initial temperature ($T_1$) = $10^\circ\text{C}$
Final temperature ($T_2$) = $-2^\circ\text{C}$
Specific heat capacity of copper ($c_{can}$) = $385 \text{ J kg}^{-1} \text{ K}^{-1}$
Specific heat capacity of water ($c_{water}$) = $4200 \text{ J kg}^{-1} \text{ K}^{-1}$
Specific heat capacity of ice ($c_{ice}$) = $2100 \text{ J kg}^{-1} \text{ K}^{-1}$
Specific latent heat of fusion of water ($L_f$) = $3.34 \times 10^5 \text{ J kg}^{-1}$

Step 1: Calculate the heat lost by the copper can as it cools from $10^\circ\text{C}$ to $-2^\circ\text{C}$.

$$Q_{can} = m_{can} c_{can} \Delta T_{can}$$ $$Q_{can} = (0.2 \text{ kg})(385 \text{ J kg}^{-1} \text{ K}^{-1})(10^\circ\text{C} - (-2^\circ\text{C}))$$ $$Q_{can} = (0.2)(385)(12) = 924 \text{ J}$$

Step 2: Calculate the heat lost by water as it cools from $10^\circ\text{C}$ to $0^\circ\text{C}$.

$$Q_{water1} = m_{water} c_{water} \Delta T_{water1}$$ $$Q_{water1} = (0.2 \text{ kg})(4200 \text{ J kg}^{-1} \text{ K}^{-1})(10^\circ\text{C} - 0^\circ\text{C})$$ $$Q_{water1} = (0.2)(4200)(10) = 8400 \text{ J}$$

Step 3: Calculate the heat lost by water as it freezes at $0^\circ\text{C}$.

$$Q_{freeze} = m_{water} L_f$$ $$Q_{freeze} = (0.2 \text{ kg})(3.34 \times 10^5 \text{ J kg}^{-1})$$ $$Q_{freeze} = 66800 \text{ J}$$

Step 4: Calculate the heat lost by the ice as it cools from $0^\circ\text{C}$ to $-2^\circ\text{C}$.

$$Q_{ice} = m_{water} c_{ice} \Delta T_{ice}$$ $$Q_{ice} = (0.2 \text{ kg})(2100 \text{ J kg}^{-1} \text{ K}^{-1})(0^\circ\text{C} - (-2^\circ\text{C}))$$ $$Q_{ice} = (0.2)(2100)(2) = 840 \text{ J}$$

Step 5: Calculate the total quantity of heat given out.

$$Q_{total} = Q_{can} + Q_{water1} + Q_{freeze} + Q_{ice}$$ $$Q_{total} = 924 \text{ J} + 8400 \text{ J} + 66800 \text{ J} + 840 \text{ J}$$ $$Q_{total} = 76964 \text{ J}$$

The total quantity of heat given out is $\boxed{\text{76964 J}}$.