Question

Water enters a pipe at \(20^{\circ} \mathrm{C}\) at a rate of $0.25 \mathrm{~kg} / \mathrm{s}\( and is heated to \)60^{\circ} \mathrm{C}$. The rate of heat transfer to the water is (a) \(10 \mathrm{~kW}\) (b) \(20.9 \mathrm{~kW}\) (c) \(41.8 \mathrm{~kW}\) (d) \(62.7 \mathrm{~kW}\) (e) \(167.2 \mathrm{~kW}\)

Short answer

Answer: The rate of heat transfer to the water is 41.8 kW.

Step-by-step solution

Identify the given values

In this problem, we are given: - Initial temperature: \(T_{1} = 20^{\circ} \mathrm{C}\) - Final temperature: \(T_{2} = 60^{\circ} \mathrm{C}\) - Mass flow rate: \(m = 0.25 \mathrm{~kg/s}\) - Specific heat capacity of water: \(c = 4.18 \mathrm{~kJ/kg^{\circ} C}\) (constant)

Calculate the temperature change

The temperature change can be calculated by taking the difference between the final and initial temperatures: \(\Delta{T} = T_{2} - T_{1} = 60^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 40^{\circ} \mathrm{C}\)

Calculate the energy transfer rate

To find the rate of heat transfer, first find the amount of energy transferred by using the formula: \(Q = m \times c \times \Delta{T}\) Now we substitute the given values into the formula: \(Q = (0.25 \mathrm{~kg/s}) \times (4.18 \mathrm{~kJ/kg^{\circ} C}) \times (40^{\circ} \mathrm{C})\)

Solve for the rate of heat transfer

Now we can calculate the rate of heat transfer: \(Q = 0.25 \times 4.18 \times 40\) \(Q = 41.8 \mathrm{~kJ/s}\) Since 1 kJ/s is equivalent to 1 kW, the rate of heat transfer to the water is \(41.8 \mathrm{~kW}\). Therefore, the correct answer is (c) \(41.8 \mathrm{~kW}\).