Question

A 2-kW electric resistance heater submerged in 30-kg water is turned on and kept on for \(10 \mathrm{~min}\). During the process, \(500 \mathrm{~kJ}\) of heat is lost from the water. The temperature rise of water is (a) \(5.6^{\circ} \mathrm{C}\) (b) \(9.6^{\circ} \mathrm{C}\) (c) \(13.6^{\circ} \mathrm{C}\) (d) \(23.3^{\circ} \mathrm{C}\) (e) \(42.5^{\circ} \mathrm{C}\)

Short answer

Answer: (a) 5.6°C

Step-by-step solution

Convert time to seconds.

Since the heater is turned on for 10 minutes, we need to convert the time to seconds for further calculations. 1 minute equals 60 seconds, so 10 minutes equals $$10 \times 60 = 600 \mathrm{~s}$$.

Calculate the energy provided by the heater.

The energy provided by the heater can be calculated using the formula $$E = Pt$$, where E is the energy, P is the power of the heater, and t is the time the heater is on. We have $$P = 2 \mathrm{~kW} = 2000 \mathrm{~W}$$ and $$t = 600 \mathrm{~s}$$. Therefore, the energy provided by the heater is $$E = 2000 \mathrm{~W} \times 600 \mathrm{~s} = 1200 \mathrm{~kJ}$$.

Calculate the net energy absorbed by the water.

We're given that 500 kJ of heat is lost during the process. So, the net energy absorbed by the water can be found by subtracting the heat loss from the energy provided by the heater: $$E_{\mathrm{net}} = 1200 \mathrm{~kJ} - 500 \mathrm{~kJ} = 700 \mathrm{~kJ}$$.

Calculate the temperature rise of water.

The formula to calculate the temperature change with the energy absorbed and the mass of water is $$\Delta T = \frac{E_{\mathrm{net}}}{mc_{\mathrm{p}}}$$, where \(\Delta T\) is the temperature change, E\(_{\mathrm{net}}\) is the net energy absorbed, m is the mass of the water, and \(c_{\mathrm{p}}\) is the specific heat capacity of the water. For water, $$c_{\mathrm{p}} = 4.18 \mathrm{~kJ/kg}^{\circ} \mathrm{C}$$ and $$m = 30 \mathrm{~kg}$$. Therefore, the temperature rise of water is $$\Delta T = \frac{700 \mathrm{~kJ}}{30 \mathrm{~kg} \times 4.18 \mathrm{~kJ/kg}^{\circ} \mathrm{C}} = 5.58^{\circ} \mathrm{C}$$. The temperature rise of water is approximately \(5.6^{\circ} \mathrm{C}\), which corresponds to the answer (a).

Key concepts

Heat Energy Transfer

Understanding how heat energy moves from one object to another is fundamental in physics and is crucial for solving problems like the one presented. Heat energy transfer can occur through three primary methods: conduction, convection, and radiation. In our exercise, we focus on the concept of heat transfer through conduction, as the electric resistance heater directly interacts with the water.

During the operation of the heater, it converts electrical energy into heat energy due to its resistance. This heat is then transferred to the water. However, not all of the heat is retained; some of it is inevitably lost to the surroundings. That's where the importance of calculating the net energy absorbed by the water comes into play. The exercise highlights that 500 kJ of heat escapes, which must be accounted for to find the actual temperature increase of the water. When solving similar problems, be mindful of the surrounding environment, which can affect the net heat transfer.

Specific Heat Capacity

The term 'specific heat capacity' (\(c_p\)) might seem daunting, but it's simply a property that indicates how much heat energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius. It's like a 'thermal bank account' for a material, showing the heat 'savings' required for a temperature 'transaction.'

The correct understanding of specific heat capacity is vital for calculating temperature changes, as seen in our exercise. Water, with a specific heat capacity of 4.18 kJ/kg°C, can 'hold' a considerable amount of heat before its temperature changes significantly. This inherent trait of water, or any other substance's specific heat capacity, must be factored into our temperature rise calculation. If unsure about the specific heat capacity values, you can generally find them in physics textbooks or reliable scientific databases.

Electric Resistance Heater

An electric resistance heater is a common device in various heating applications, both practical and experimental. By passing an electric current through a resistive material, it utilizes the principle of Joule heating; the resistance to the current flow generates heat. This is often seen in household items like toasters or electric stoves.

In our exercise, the power rating of 2 kW indicates the rate at which the heater uses energy to produce heat. Tying it into our earlier concepts, the heat energy transfer from this heater to the water determines the rise in temperature. It's crucial to convert the power to a common unit of energy (joules or kilojoules) and to account for time to find out how much energy is actually transferred. This is a good example of practical thermodynamics and gives a clear view of energy transformation in everyday appliances.