Thermal physics problem-solving question

A student investigates energy transfer between a hot metal block and a liquid. An aluminium block of mass $0.200\,\mathrm{kg}$ is heated from $20.0\,^{\circ}\mathrm{C}$ to $100.0\,^{\circ}\mathrm{C}$. The heater is then removed. The dry block is placed completely into a cup containing $0.100\,\mathrm{kg}$ of water, initially at $20.0\,^{\circ}\mathrm{C}$. The water is stirred gently until the block and water reach the same temperature. Use the following data: specific heat capacity of aluminium: $900\,\mathrm{J\,kg^{-1}\,K^{-1}}$ specific heat capacity of water: $4200\,\mathrm{J\,kg^{-1}\,K^{-1}}$ For parts (a) and (b), assume that the specific heat capacities are constant, no energy is lost to the surroundings, and the cup, thermometer and stirrer absorb negligible energy. The block is still at $100.0\,^{\circ}\mathrm{C}$ when it enters the water. No material changes state and no water is lost. (a) Calculate the energy transferred to the aluminium block as it is heated from $20.0\,^{\circ}\mathrm{C}$ to $100.0\,^{\circ}\mathrm{C}$. [2] (b) Calculate the common final temperature of the aluminium block and the water. Show how conservation of energy is used in your calculation. [4] (c) In the real experiment, the room is at $20.0\,^{\circ}\mathrm{C}$. Some energy is transferred to the surroundings while the hot block is moved and while the water is stirred. Continue to neglect energy absorbed by the cup, thermometer and stirrer. State whether the temperature recorded when the block and water first reach the same temperature will be higher than, lower than, or equal to your answer to (b). Explain your answer in terms of energy transfer. [2] (d) Suggest one practical change that would reduce energy transfer to the surroundings during this experiment. Explain how your change would help. [2]

(a) The temperature change of the block is $$\Delta\theta=100.0-20.0=80.0\,\mathrm{K}.$$ Using $E=mc\Delta\theta$: $$E=(0.200)(900)(80.0)=1.44\times10^4\,\mathrm{J}.$$ (b) Let the common final temperature be $\theta$ in degrees Celsius. The block cools from $100.0\,^{\circ}\mathrm{C}$ to $\theta$, while the water warms from $20.0\,^{\circ}\mathrm{C}$ to $\theta$. With no energy transferred elsewhere, energy lost by the aluminium equals energy gained by the water: $$m_{\mathrm{Al}}c_{\mathrm{Al}}(100.0-\theta)=m_{\mathrm{w}}c_{\mathrm{w}}(\theta-20.0).$$ $$(0.200)(900)(100.0-\theta)=(0.100)(4200)(\theta-20.0).$$ $$180(100.0-\theta)=420(\theta-20.0).$$ $$18000-180\theta=420\theta-8400,$$ $$26400=600\theta,$$ $$\boxed{\theta=44.0\,^{\circ}\mathrm{C}}.$$ Check: the aluminium loses $(0.200)(900)(100.0-44.0)=10080\,\mathrm{J}$. The water gains $(0.100)(4200)(44.0-20.0)=10080\,\mathrm{J}$. The block retains $4320\,\mathrm{J}$ of the energy supplied in (a), relative to its initial temperature. Therefore the whole $14400\,\mathrm{J}$ from (a) is not transferred to the water. (c) The recorded common temperature will be lower than $44.0\,^{\circ}\mathrm{C}$. Some of the energy supplied to the block is transferred to the cooler surroundings instead of remaining in the block-water system. Less energy is available to raise the water's temperature, so the system reaches a lower common temperature. (d) For example, put insulating material around the cup. This reduces energy transfer through the cup walls to the cooler surroundings during mixing. Other valid changes include fitting a lid...

(a) [2] M1: Uses E = mc delta theta with a temperature change of 80.0 K (or 80.0 degrees C), mass 0.200 kg and c = 900 J kg^-1 K^-1. [1] A1: Obtains 14400 J or 14.4 kJ, with a correct unit. [1] (b) [4] M1: Applies conservation of energy: energy lost by aluminium equals energy gained by water, or an equivalent total-energy balance with a common reference temperature. [1] M1: Uses the two correct temperature changes, 100.0 - theta for the aluminium and theta - 20.0 for the water, with the correct heat-capacity factors: 180(100.0 - theta) = 420(theta - 20.0). [1] A1: Correctly rearranges the balance, for example 26400 = 600 theta, or correctly evaluates the equivalent heat-capacity-weighted mean. [1] A1: Obtains a common temperature of 44.0 degrees C (accept 44 degrees C), or 317.15 K (accept 317 K when rounded to three significant figures), with a unit. [1] Accept fully consistent Kelvin working. Correct conservation-based working can earn method marks after a subsequent arithmetic...

For (a), use the change in temperature rather than the final temperature. At the common final temperature, energy lost by the aluminium equals energy gained by the water under the ideal assumptions. Call the final temperature theta. The aluminium cools through 100.0 - theta; the water warms through theta - 20.0. For (c) and (d), identify where energy goes and how a change reduces that transfer.