/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Water flows at the rate of \(0.1... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 65

Water flows at the rate of \(0.1500 \mathrm{~kg} / \mathrm{min}\) through a tube and is heated by a heater dissipating \(25.2 \mathrm{~W}\). The inflow and outflow water temperatures are \(15.2^{\circ} \mathrm{C}\) and \(17.4^{\circ} \mathrm{C}\), respectively. When the rate of flow is increased to \(0.2318 \mathrm{~kg} / \mathrm{min}\) and the rate of heating to \(37.8 \mathrm{~W}\), the inflow and outflow temperatures are unaltered. Find: (A) The specific heat capacity of water. (B) The rate of loss of heat from the tube.

Short Answer

Expert verified
(A) The specific heat capacity of water is \(70 \, \mathrm{J/kg^{\circ}C}\). (B) The rate of loss of heat from the tube is \(2.1 \, \mathrm{W}\).

Step by step solution

01

Set up the equations in terms of specific heat capacity and heat loss

We are given the flow rates, heating rates, and the inflow and outflow temperatures for two different scenarios. Let's assume that the specific heat capacity of water is \(c\) and the rate of heat loss from the tube is \(L\). The equations for both scenarios can be set up as follows: For scenario 1 (Flow rate = 0.1500 kg/min, Heating rate = 25.2 W): \( 0.1500c(17.4 - 15.2) = 25.2 - L \) For scenario 2 (Flow rate = 0.2318 kg/min, Heating rate = 37.8 W): \( 0.2318c(17.4 - 15.2) = 37.8 - L \)
02

Solve the equations simultaneously

To find \(c\) and \(L\), we need to solve these equations simultaneously. We can use the elimination method to do this. First, let's get rid of the L variable from both equations by subtracting the second equation from the first: \(0.1500c(17.4 - 15.2) - 0.2318c(17.4 - 15.2) = 25.2 - 37.8 \)
03

Simplify the equation and solve for specific heat capacity

Now, we have an equation with only one variable, \(c\). Let's simplify the equation and solve for the specific heat capacity. \( -0.0818c(2.2) = -12.6 \) \( -0.180c = -12.6 \) \( c = \frac{-12.6}{-0.180} \) \( c = 70 \, \mathrm{J/kg^{\circ}C} \) So, the specific heat capacity of water is 70 J/kg掳C.
04

Calculate the rate of heat loss from the tube

Now that we have the specific heat capacity, we can substitute it back into either equation from Step 1 to find the rate of heat loss (\(L\)). We will use the equation for scenario 1: \( 0.1500 \cdot 70 (17.4 - 15.2) = 25.2 - L \) \( 0.1500 \cdot 70 (2.2) = 25.2 - L \) \( 23.1 = 25.2 - L \) \( L = 25.2 - 23.1 \) \( L = 2.1 \, \mathrm{W} \) Thus, the rate of loss of heat from the tube is 2.1 W. Answers: (A) The specific heat capacity of water is 70 J/kg掳C. (B) The rate of loss of heat from the tube is 2.1 W.

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Most popular questions from this chapter

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