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Question
(a) How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from $30.6^\circ\textrm{C}$ to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W? 1 watt = 1 joule / second (1W = 1 J/s).
1. $2.47 \times 10^6 \textrm{ J}$
2. $4.94 \times 10^3 \textrm{ s}$
Solution Video

# OpenStax College Physics Solution, Chapter 14, Problem 13 (Problems & Exercises) (2:40)

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Video Transcript
This is College Physics Answers with Shaun Dychko. In this question we have 2.5 kilograms of water inside a 0.75 kilogram of aluminum pot and water has an initial temperature of 30 degree Celsius and we told that its temperature will be raised to its boiling point which is 100 degree Celsius and then 0.75 kilogram will be boiled away. So, how much heat energy is required to do all that. That’s part a. And then part b says how long will it take given the power source of 500 watts. So, we have the…. We have parts of this question well three parts you might say. We have the energy required to heat the aluminum and change its temperature from 30 to 100 because the aluminum pot will be the same temperature as the water inside it and then also wanted to heat the water as well. So, these two terms are for the change of temperature of the aluminum and water from 30 to 100 and then add to that the amount of energy required to change the state of three quarters of kilogram of water and so we multiply that small three quarters of kilogram mass by the latent heat of vaporization of water. So, this change of temperature can be factor out from these two terms and we written that as T f minus T initial and then we multiplying by the mass of aluminum times specific heat of aluminum plus the same corresponding term for water and then add to that the latent heat of vaporization multiplied by the mass that’s vaporized and then plug in numbers. So, we have a 100 degree Celsius final temperature minus 30 degree Celsius initial temperature for both the aluminum and water multiplied by three quarters of kilogram of aluminum times 900 joules per kilogram per Celsius degree specific heat for aluminum plus 2.5 kilogram of water times specific heat of water and then add to that three quarters of kilogram of water that is vaporized multiplied by the latent heat of vaporization for water of 2256 times ten to the three joules per kilogram vaporized and this workout to 2.47 times ten to the six joules. Now the part b asks does how long it will take and so power is the amount of energy divided by the time a in… for that energy to be transferred and we solved for T by multiplying both sides by T over P and on left side we have T and on right side we have Q over P. so, we have the answer from part a it will energy divided by 500 watts which gives 4.94 times ten to the three seconds.

Submitted by mengar90 on Thu, 08/12/2021 - 02:27

Isn't the specific heat of water 4184 J/ kg C instead of 4186 J/kg C? I thought it was equivalent to 1 kcal

Submitted by ShaunDychko on Mon, 08/16/2021 - 16:06

Hi mengar90, thank you for the question. The value I'm using is from Table 14.1 in the OpenStax textbook. Specific heat is temperature and pressure dependent so it's normal to see different values mentioned in different sources depending on the pressure/temperature each source is assuming. The OpenStax text assumes $15^\circ\textrm{ C}$ and 1 atmosphere of pressure.

kcal is a unit of energy, and yes you're correct that it's 4184 joules, but note the units. The comparison is between two different units for energy: kcal vs. J, not kcal vs J / kg C.

Hope this helps,
Shaun