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Chapter 1: Q.1.28 (page 20)

Estimate how long it should take to bring a cup of water to boiling temperature in a typical 600-wattmicrowave oven, assuming that all the energy ends up in the water. (Assume any reasonable initial temperature for the water.) Explain why no heat is involved in this process.

Short Answer

Expert verified

Time is taken to boil2.616 minutes to raise the 20mlwater fromlocalid="1648467354265" 10°Cto100°C.

Step by step solution

01

Step 1. Finding Amount of heat.

Consider a mug filled with250mLcold10°Cwater in a 600wattmicrowave. To achieve the boiling point, So must raise the temperature by90˚C.

Q=mCΔT

m=Water mass.

C=Specific heat C=4.186J/g·C∘

ΔT=90°Temp

Therefore,

Q=250×4.186×90=94185J.

02

Step 2. Boiling time.

The Power

P=Qt→t=QP

Insert Qand Pvalues.

t=94185600=156.975s=2.616min.

Normally, heat is transferred from the hotter to the colder object. However, because there is no hot item in our scenario, the heat is transported to the mug via an electromagnetic wave generated by the magnetron.

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Most popular questions from this chapter

Problem 1.49. Consider the combustion of one mole of H2with1/2 mole ofO2 under standard conditions, as discussed in the text. How much of the heat energy produced comes from a decrease in the internal energy of the system, and how much comes from work done by the collapsing atmosphere? (Treat the volume of the liquid water as negligible.)

Imagine some helium in a cylinder with an initial volume of 1litreand an initial pressure of 1atm.Somehow the helium is made to expand to a final volume of 3litres,in such a way that its pressure rises in direct proportion to its volume.

(a) Sketch a graph of pressure vs. volume for this process.

(b) Calculate the work done on the gas during this process, assuming that there are no "other" types of work being done.

(c) Calculate the change in the helium's energy content during this process.

(d) Calculate the amount of heat added to or removed from the helium during this process.

(e) Describe what you might do to cause the pressure to rise as the helium expands.

Calculate the rate of heat conduction through a layer of still air that is1mmthick, with an area of 1m2, for a temperature difference of 20∘C.

When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 550,000 . The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β :
β≡ΔV/VΔT
(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β =1 / 550,000 K-1=1.81 x 10-4 K-1. (The exact value varies with temperature, but between 0oC and 200oC the variation is less than 1 %.)
(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.
(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 x 10 -4 K-1 at 100oC, but decreases as the temperature is lowered until it becomes zero at 4oC. Below 4oC it is slightly negative, reaching a value of -0.68 x 10-4K-1 at 0oC. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.

Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep Vfixed as Tincreases, as follows.

(a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume,dV1, in terms of dTand the thermal expansion coefficient βintroduced in Problem 1.7.

(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, dV2, in terms of dPand the isothermal compressibility κT, defined as

κT≡−1V∂V∂PT

(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dPtodTis equal to (∂P/∂T)V, since there is no net change in volume. Express this partial derivative in terms of βandκT. Then express it more abstractly in terms of the partial derivatives used to define βandκT. For the second expression you should obtain

∂P∂TV=−(∂V/∂T)P(∂V/∂P)T

This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.

(d) Compute β,κT,and(∂P/∂T)Vfor an ideal gas, and check that the three expressions satisfy the identity you found in part (c).

(e) For water at 25∘C,β=2.57×10−4K−1andκT=4.52×10−10Pa−1. Suppose you increase the temperature of some water from 20∘Cto30∘C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at25∘C)β=1.81×10−4K−1andκT=4.04×10−11Pa−1

Given the choice, would you rather measure the heat capacities of these substances at constant vor at constant p?
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