The water in a small wildlife rescue tank is warmed by an immersion heater after a power cut. Fig. 1 shows the heater connected to a supply. A data logger r...
Question 1 Report
The water in a small wildlife rescue tank is warmed by an immersion heater after a power cut. Fig. 1 shows the heater connected to a supply. A data logger records the temperature of 1.20 kg of water. The heater is used with a 120 V supply and takes a current of 4.5 A. It is switched on for 180 s. The water temperature rises from 16.0 degrees C to 31.0 degrees C. Some energy heats the glass tank and some is transferred to the room by radiation and convection.
(a) What is the equation linking electrical power, potential difference and current? [1]
(b) What is the power of the heater? [2]
(c) What electrical energy is supplied to the heater during the 180 s test? [2]
(d) Which calculation gives the thermal energy gained by the water? Use a specific heat capacity of water of 4200 J / kg degrees C. [3]
(e) Sketch, in words, one reason why the electrical energy supplied is greater than the energy gained by the water. [2]
Answer Details
(a) \[P=VI\] [1]
(b) \[P=120\times4.5=540\text{ W}\] [2]
(c) Use \(E=Pt\):
\[E=540\times180=97\,200\text{ J}\] [2]
(d) First find the temperature rise:
\[\Delta T=31.0-16.0=15.0^\circ\text{C}\]
Then use \(E=mc\Delta T\):
\[E=1.20\times4200\times15.0=75\,600\text{ J}\] [3]
(e) The electrical input is greater because some energy heats the glass tank rather than the water, and some is transferred from the heater or tank to the room by radiation or convection. [2]
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