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Chapter 3: Problem 192

A \(1.2-\mathrm{kg}\) pot of water at \(20^{\circ} \mathrm{C}\) is put on a stove supplying \(250 \mathrm{~W}\) to the water. How long will it take to come to a boil \(\left(100^{\circ} \mathrm{C}\right) ?\)

Short Answer

Expert verified
It will take approximately 26.8 minutes for the water to come to a boil.

Step by step solution

01

Identify Given Information

We have a pot of water with a mass of \(1.2\, \mathrm{kg}\) initially at a temperature of \(20^{\circ}\, \mathrm{C}\). The specific heat capacity of water is approximately \(4.186\, \mathrm{J/g^{\circ}C}\) or \(4186\, \mathrm{J/kg^{\circ}C}\). The power supplied by the stove is \(250\, \mathrm{W}\). We need to find the time it takes for the water to heat up to \(100^{\circ}\, \mathrm{C}\).
02

Calculate the Temperature Change

The initial temperature is \(20^{\circ}\, \mathrm{C}\) and the final temperature is \(100^{\circ}\, \mathrm{C}\). Therefore, the change in temperature \(\Delta T\) is \(100 - 20 = 80^{\circ}\, \mathrm{C}\).
03

Calculate the Heat Required

Use the formula for heat added: \(Q = mc\Delta T\), where \(m = 1.2\, \mathrm{kg}\), \(c = 4186\, \mathrm{J/kg^{\circ}C}\), and \(\Delta T = 80^{\circ}\, \mathrm{C}\). Substitute these into the formula to find \(Q\).\[Q = (1.2\, \mathrm{kg})(4186\, \mathrm{J/kg^{\circ}C})(80^{\circ}\, \mathrm{C})\]\[Q = 401856\, \mathrm{J}\]
04

Find the Time Required Using Power

Power \(P\) is defined as the energy per unit time, \(P = \frac{Q}{t}\). Rearrange this equation to solve for time \(t\):\[t = \frac{Q}{P}\]Substitute the known values of \(Q = 401856\, \mathrm{J}\) and \(P = 250\, \mathrm{W}\):\[t = \frac{401856}{250}\]\[t = 1607.424\, \mathrm{seconds}\]
05

Convert Time to Minutes

Convert the time from seconds to minutes by dividing by 60:\[\frac{1607.424}{60} \approx 26.79\, \mathrm{minutes}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
Specific heat capacity is a crucial concept in understanding how substances absorb heat. It is the amount of heat required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius. This property varies from one material to another, influencing how quickly or slowly it heats up.
For water, a common material, the specific heat capacity is quite high at approximately 4186 J/kg掳C. This means water can absorb a lot of heat without a significant change in temperature, making it ideal for regulating heat in various processes. Knowing the specific heat capacity helps us calculate how much energy is necessary to change the temperature of a given mass of water.
Temperature Change
Temperature change in thermodynamics refers to the difference between the final and initial temperatures of a substance. It is often denoted as \(\Delta T\).
In problems involving heating, like boiling water, identifying the temperature change is vital to determining the heat required. For instance, if water is initially at 20掳C and needs to reach 100掳C to boil, the temperature change \(\Delta T\) is 80掳C. Recognizing \(\Delta T\) allows us to use it in the heat required calculation effectively.
Heat Required
Calculating the heat required to change a substance's temperature involves applying the formula \(Q = mc\Delta T\).
  • \(Q\) is the heat energy in joules.
  • \(m\) is the mass of the substance, in kilograms.
  • \(c\) is the specific heat capacity.
  • \(\Delta T\) is the temperature change.
For example, to boil 1.2 kg of water from 20掳C to 100掳C, the heat required (Q) is calculated using the specific heat capacity of water (4186 J/kg掳C). By substituting the values into the formula, we find that \(Q\) is 401856 joules. This calculation tells us the total amount of energy needed for the water to reach boiling point.
Power and Energy Relationship
Understanding the relationship between power and energy sheds light on how quickly energy can do work. Power (P) is the rate at which energy is used or transferred, expressed in watts (W), where 1 W equals 1 joule per second.
To find how long it takes to heat the water, we use the equation \(P = \frac{Q}{t}\), where Q is the energy in joules and \(t\) is the time in seconds. Rearranging the equation to solve for time gives \(t = \frac{Q}{P}\).
In our example, with 250 W power supplied, the time needed is \(t = \frac{401856}{250}\), resulting in roughly 1607 seconds. Converting seconds into minutes, we understand the water takes approximately 26.79 minutes to come to a boil.

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Most popular questions from this chapter

A hydraulic hoist raises a 3650 -lbm car \(6 \mathrm{ft}\) in an auto repair shop. The hydraulic pump has a constant pressure of \(100 \mathrm{lbf} / \mathrm{in} .^{2}\) on its piston. What is the increase in potential energy of the car, and how much volume should the pump displace to deliver that amount of work?

A water heater is covered with insulation boards over a total surface area of \(3 \mathrm{~m}^{2}\). The inside board surface is at \(75^{\circ} \mathrm{C},\) the outside surface is at \(18^{\circ} \mathrm{C},\) and the board material has a conductivity of \(0.08 \mathrm{~W} / \mathrm{m} \mathrm{K}\). How thick should the board be to limit the heat transfer loss to \(200 \mathrm{~W} ?\)

A large condenser (heat exchanger) in a power plant must transfer a total of \(100 \mathrm{MW}\) from steam running in a pipe to seawater being pumped through the heat exchanger. Assume that the wall separating the steam and seawater is \(4 \mathrm{~mm}\) of steel, with conductivity of \(15 \mathrm{~W} / \mathrm{m} \mathrm{K},\) and that a maximum \(5^{\circ} \mathrm{C}\) difference between the two fluids is allowed in the design. Find the required minimum area for the heat transfer, neglecting any convective heat transfer in the flows.

The brake shoe and steel drum of a car continuously absorb \(75 \mathrm{~W}\) as the car slows down. Assume a total outside surface area of \(0.1 \mathrm{~m}^{2}\) with a convective heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) to the air at \(20^{\circ} \mathrm{C}\). How hot does the outside brake and drum surface become when steady conditions are reached?

The temperature of water at \(400 \mathrm{kPa}\) is raised from \(150^{\circ} \mathrm{C}\) to \(1200^{\circ} \mathrm{C}\). Evaluate the change in specific internal energy using (a) the steam tables, (b) the ideal gas Table \(\mathrm{A} .8\), and (c) the specific heat Table A.5.
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