Time / s Temperature / °C 0 45 30 38 60 30 90 23 120 17 150 13 180 12 210 12 The apparatus in Fig. 1.1 is set up to investigate how the temperature of warm ...
Question 1 Report
| Time / s | Temperature / °C |
|---|---|
| 0 | 45 |
| 30 | 38 |
| 60 | 30 |
| 90 | 23 |
| 120 | 17 |
| 150 | 13 |
| 180 | 12 |
| 210 | 12 |
The apparatus in Fig. 1.1 is set up to investigate how the temperature of warm water changes when ice cubes are added. A beaker contains 0.40 kg of water initially at 45 °C. Several ice cubes at 0 °C are placed in the water and the mixture is stirred continuously. A thermometer records the temperature every 30 seconds.
Table 1.1 shows the temperature readings.
(a) Record the temperature at time 0 from Table 1.1. [1]
(b) Plot a graph of temperature (vertical axis) against time (horizontal axis) using the data in Table 1.1. [2]
(c) From your graph, state the time at which the temperature stops falling. [1]
(d) Calculate the total temperature drop from the start until the reading stabilises. [1]
(e) Calculate the energy lost by the water as it cools from 45 °C to 12 °C. Use the equation energy = mass x specific heat capacity x temperature change. Take c = 4200 J/(kg °C). Show your working. [2]
(f) Explain, in terms of energy and particles, where this lost energy goes. [2]
(g) Suggest why the student stirs the mixture throughout the experiment. [1]
Answer Details
(a) The temperature at time 0 is 45 °C. [1]
This is read directly from the first row of Table 1.1.
(b) The correctly plotted graph is shown below, with temperature on the vertical axis and time on the horizontal axis. [2]
The data points are plotted and joined with straight lines. The graph shows a steep initial decline that gradually flattens out as the mixture approaches thermal equilibrium.
(c) From the graph, the temperature stops falling at approximately 180 s (accept 150 to 180 s). [1]
After about 180 s the graph becomes horizontal, indicating that the water temperature is no longer changing. The mixture has reached thermal equilibrium with the melted ice.
(d) Total temperature drop:
\[ \Delta\theta = 45 - 12 = 33 \; ^\circ\text{C} \][1]
(e) Using the equation \( E = mc\Delta\theta \):
\[ E = 0.40 \times 4200 \times 33 \; [1] \] \[ E = 55\,440 \; \text{J} \approx 55\,000 \; \text{J} \; [1] \]The mass of the water is 0.40 kg, the specific heat capacity of water is 4200 J/(kg °C), and the temperature change is 33 °C.
(f) The thermal energy lost by the warm water is absorbed by the ice cubes. Part of this energy breaks the bonds between ice particles, melting the ice (this is the latent heat of fusion). [1] The remaining energy warms the resulting meltwater from 0 °C up to the final equilibrium temperature of 12 °C. [1]
Energy is conserved: the energy lost by the warm water equals the energy gained by the ice (for melting) plus the energy gained by the meltwater (for warming).
(g) Stirring ensures the water temperature is uniform throughout the beaker, so the thermometer gives an accurate and representative reading at each time interval. [1]
Without stirring, cooler water would collect near the ice while warmer water remained elsewhere. The thermometer reading would depend on its position rather than reflecting the true average temperature of the mixture.
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