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Chapter 8: Problem 1111

Water of volume 2 liter in a container is heated with a coil of $1 \mathrm{kw}\( at \)27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at the rate of \(160(\mathrm{~J} / \mathrm{S}) .\) In how much time temperature will rise from \(27^{\circ} \mathrm{C}\) to $77^{\circ} \mathrm{C}\(. Specific heat of water is \)4.2\\{(\mathrm{KJ}) /(\mathrm{Kg})\\}$ (A) \(7 \mathrm{~min}\) (B) \(6 \min 2 \mathrm{~s}\) (C) \(14 \mathrm{~min}\) (D) \(8 \min 20 \mathrm{~S}\)

Short Answer

Expert verified
The time required for the water's temperature to rise from $27^{\circ}\mathrm{C}$ to $77^{\circ}\mathrm{C}$ is \(8 \) minutes and \(20 \) seconds (D).

Step by step solution

01

Convert volumes and temperatures to the appropriate units.

First, we need to convert the volume of water from liters to kilograms, as we know the specific heat capacity of water in (kJ/kg)(K). We also need to convert the power of the coil from kW to J/s. Water has a density of 1 kg/L, so: 2 L × 1 kg/L = 2 kg 1 kW = 1000 J/s Now, we have: - Mass of water (m) = 2 kg - Initial temperature (Ti) = 27°C - Final temperature (Tf) = 77°C - Power of the coil (Pc) = 1000 J/s - Energy dissipation rate (Ed) = 160 J/s - Specific heat of water (Cw) = 4.2 kJ/kg(K) = 4200 J/kg(K)
02

Find the net power supplied to the water.

To find the net power supplied to the water, we need to account for the power provided by the coil and the energy dissipation. We'll subtract the energy dissipation rate from the power of the coil: Net power (Pn) = Pc - Ed Pn = 1000 J/s - 160 J/s Pn = 840 J/s
03

Calculate the heat required to increase the temperature of water.

In order to determine the heat required (ΔQ) to increase the temperature from Ti to Tf, we'll use the formula: ΔQ = mcΔT where: m = mass of water c = specific heat of water ΔT = temperature difference ΔT = Tf - Ti = 77°C - 27°C = 50°C Now, calculate the heat required: ΔQ = 2 kg × 4200 J/kg(K) × 50 K ΔQ = 420,000 J
04

Calculate the time required to raise the temperature.

Now we'll use the net power (Pn) and the heat required (ΔQ) to calculate the time (t) it takes for the temperature to increase: t = ΔQ / Pn t = 420,000 J / 840 J/s t = 500 seconds
05

Convert time to minutes and seconds and find the answer.

To convert the time to minutes and seconds, we'll use basic time conversion rules: 1 minute = 60 seconds 500 seconds = 8 minutes and 20 seconds The answer is (D) 8 minutes 20 seconds.

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Most popular questions from this chapter

Instructions:Read the assertion and reason carefully to mask the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not be correct explanation of assertion. (C) If assertion is true but reason is false. (D) If the assertion and reason both are false. Assertion: The carnot is useful in understanding the performance of heat engine Reason: The carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of given temperatures. (A) \(\mathrm{A}\) (B) B (C) \(\mathrm{C}\) (D) \(\mathrm{D}\)

An ideal gas at \(27 \mathrm{C}\) is Compressed adiabatically, to \(\\{8 / 27\\}\) of its original Volume. If \(\mathrm{v}=(5 / 3)\), then the rise in temperature is (A) \(225 \mathrm{k}\) (B) \(450 \mathrm{~K}\) (C) \(375 \mathrm{~K}\) (D) \(405 \mathrm{~K}\)

If du represents the increase in internal energy of a thermodynamic system and dw the work done by the system, which of the following statement is true? (A) \(\mathrm{du}=\mathrm{dw}\) in isothermal process (C) \(\mathrm{du}=-\mathrm{dw}\) in an aidabatic process (B) \(\mathrm{du}=\mathrm{dw}\) in aidabatic process (D) \(\mathrm{du}=-\mathrm{dw}\) in an isothermal process

In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas release \(20 \mathrm{~J}\) of heat and $8 \mathrm{~J}$ of work has done on the gas-If the initial internal energy of the gas was \(30 \mathrm{j}\), then the final internal energy will be (A) \(58 \mathrm{~J}\) (B) \(2 \mathrm{~J}\) (C) \(42 \mathrm{~J}\) (D) \(18 \mathrm{~J}\)

One mole of a monoatomic gas is heat at a constant pressure of 1 atmosphere from \(0 \mathrm{k}\) to \(100 \mathrm{k}\). If the gas constant $\mathrm{R}=8.32 \mathrm{~J} / \mathrm{mol} \mathrm{k}$ the change in internal energy of the gas is approximate ? (A) \(23 \mathrm{~J}\) (B) \(1.25 \times 10^{3} \mathrm{~J}\) (C) \(8.67 \times 10^{3} \mathrm{~J}\) (D) \(46 \mathrm{~J}\)
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