Question

A 2 -kW electric resistance heater submerged in \(5-\mathrm{kg}\) water is turned on and kept on for 10 min. During the process, \(300 \mathrm{kJ}\) of heat is lost from the water. The temperature rise of water is \((a) 0.4^{\circ} \mathrm{C}\) (b) \(43.1^{\circ} \mathrm{C}\) \((c) 57.4^{\circ} \mathrm{C}\) \((d) 71.8^{\circ} \mathrm{C}\) \((e) 180^{\circ} \mathrm{C}\)

Short answer

a) 34.9°C b) 43.1°C c) 55.0°C d) 66.3°C Answer: b) 43.1°C

Step-by-step solution

Calculate the energy provided by the heater

First, let's determine the amount of energy the heater provides to the water. Since the power of the heater is 2 kW (2000 W) and it's turned on for 10 minutes (600 seconds), we can find the energy in Joules using: Energy = Power * Time Energy = 2000 W * 600 s Energy = 1.2 * 10^6 J

Calculate the net energy gain of water

Now, we have to subtract the heat lost from the total energy provided by the heater: Net energy gain = Energy provided - Heat lost Net energy gain = 1.2 * 10^6 J - 300 kJ Net energy gain = 1.2 * 10^6 J - 300 * 10^3 J Net energy gain = 900 * 10^3 J = 9 * 10^5 J

Find the temperature rise

Given the net energy gain of water and its mass, we can calculate the temperature rise using the specific heat capacity of water. The specific heat capacity of water is 4190 J/(kg*K). We use the formula: Temperature rise = Net energy gain / (Mass * Specific heat capacity) Temperature rise = (9 * 10^5 J) / (5 kg * 4190 J/(kg*K)) Temperature rise = (9 * 10^5) / (5 * 4190) Temperature rise = 0.0859 * 10^5 ≈ 43.1°C So, the correct answer is (b) \(43.1^{\circ} \mathrm{C}\).