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Homework Help: Converting joules to calories

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Convert 4.18 joules/(grams * Kelvin) to calories/(pound * Farenheit)

    2. Relevant equations
    1 Kelvin = -272 Celcius
    1 Calorie = 4.18 Joules
    1 Celcius = 33.8 Fahrenheit
    1 lb. = 454 grams

    3. The attempt at a solution
    [itex]\frac{4.18 Joules}{1 gram \cdot 1 Kelvin} \cdot \frac{-272 Kelvin}{1 Celcius} \cdot \frac{1 Celcius}{33.8 Fahrenheit} \cdot \frac{1 Calorie}{4.18 Joules} \cdot \frac{4.54 grams}{1 pound} \\\\
    \frac{-272 \cdot 4.54 calories}{33.8 lb. \cdot Fahrenheit} \\\\
    \frac{-36.5349 calories}{lb. \cdot Fahrenheit}[/itex]

    But the solution is 252 cal/(lb. * fahrenheit). What did I do incorrectly?
  2. jcsd
  3. Feb 3, 2013 #2


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    When converting temperature units, all that matters is the size of one unit relative to another. The fact that the different temperature scales have different zero-points is irrelevant. From the point of view of unit conversion, 1 K = 1°C, since those units have the same SIZE. A step in temperature by 1 K is the same as a step in temperature by 1°C. Similarly, 1°C = (9/5)°F, since these units have different sizes. A step in temperature by 1°C is bigger than a step in temperature by 1°F. Bigger by a factor of 9/5.
    Last edited: Feb 3, 2013
  4. Feb 3, 2013 #3


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    If 1 lb = 454 g, then why did you write 4.54 g/lb?
  5. Feb 3, 2013 #4
    When converting from Kelvin to Celsius, each 1 Kelvin is in fact 1 Celsius, so you didn't need the -272. It's not asking what the Kelvin is at 1 Celsius, it's pretty much asking for every one Celsius how many Kelvin, and the answer to that would be 1.
  6. Feb 4, 2013 #5
    Okay, so here is my revised work:

    [itex]\frac{4.18 Joules}{1 gram \cdot 1 Kelvin} \cdot \frac{1 Kelvin}{1 Celcius} \cdot \frac{1 Celcius}{\frac{9}{5} Fahrenheit} \cdot \frac{1 Calorie}{4.18 Joules} \cdot \frac{454 grams}{1 pound} \\\\
    \frac{454 calories}{\frac{9}{5} \cdot Fahrenheit} \\\\
    \frac{252 calories}{lb. \cdot Fahrenheit}[/itex]

    Looks like it works out then.

    I still don't really understand how every 1 celcius is 1 kelvin. For now, all I can think of is like a linear function K = 273 + 1C where 1C is the slope, and F = (9/5)C + 32 where (9/5) is the slope. Not sure how to get the explanation that for every increase in 1 kelvin, there is an increase of 1 celcius or something like that. I don't understand how temperature is a size...I realize that it's a scalar quantity, but other than that..
  7. Feb 4, 2013 #6


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    That's precisely the point. The slope is all you care about for the unit conversion. The slope tells you the rate at which one quantity changes with the other one. If you know calculus, then what we're saying is that dK/dC = d/dC(273 + 1*C) = 1. Perhaps fanofdean's wording was more helpful than mine. How many degrees Celcius are there for every one Kelvin? Ans: ONE.

    Another way to explain it:

    If the temperature outside increases by 1°C, did it change by 273 kelvins? Absolutely not. Otherwise we'd all be dead. The temperature changed by 1 kelvin. So it's totally wrong in unit conversion to say that there is one degree celsius for every 273 kelvins. There isn't. There is one degree Celsius for every one kelvin. EDIT: You were confused by how a temperature unit can be a size. When I say the units have the same "size", I mean that they correspond the same increment in temperature.

    We're not talking about values of "the temperature," which depend on an arbitrary reference point. Obviously if T = +1°C, then T = 274 K. However, we're talking about sizes of the unit. So the statement about sizes of units is:

    1 Celsius degree = 1 kelvin (comparison of unit sizes)


    1 Celsius degree = 1.8 Fahrenheit degrees (comparison of unit sizes)

    As opposed to a statement about the temperature outside, which is

    T = 1 degree Celsius = 274 kelvins (measurement of temperature relative to arbitrary reference point that has two different values on two different scales)

    What you're REALLY saying with the latter statement is:


    The difference is that the freezing point of water is set to 0 on the Celsius scale, whereas the freezing point of water is set to 273 on the kelvin scale, so we end up with

    T = 1°C + 0°C = 1 K + 273 K. The only difference is in reference point.


    Let's say we have two different units for measuring altitude. One is the metre. On the standard scale used by scientists, the metre is the unit of altitude, h, and the reference point for h is set to sea level. I.e. h is defined to be 0 m at sea level on this altitude scale.

    However, suppose in a particular country called Kanadia, for civil (non-scientific) altitude measurement purposes, a totally different altitude measurement scale is used. The unit of altitude is called the kanad. Furthermore, altitude is measured relative to elevation of Kanadia's largest city, which is 1000 m above sea level. So h = 0 kanads corresponds to h = 1000 m. So, when an airplane's altitude is h = 9473 m on the scientific scale, the Kanadian TV stations would report it as being at h = 8473 kanads. The thing is, if you were to take a "kanad-stick", which is a stick of length one kanad, and compare side by side to to a metre stick, you would find that they are exactly the same length. In terms of the sizes of the units, 1 kanad = 1 metre. So, when doing a unit conversion from kanads to metres, you essentially need to find the answer to the question, "how many kanads are there for every one metre?" The answer is ONE. The conversion factor is ONE. Not 1000. The two altitude scales have different zero points, but the units of measurement are the same size for both.

    EDIT: I put the really crucial thing in boldface. I think it gets to the crux of your confusion. As you can see, the value of T relative to the reference point is "+one unit" in both cases.
    Last edited: Feb 4, 2013
  8. Feb 4, 2013 #7


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    Here, it's useful to understand the equations from which the quantities come. The quantity 4.18 joules/(grams * Kelvin) is the heat capacity of water (C) and is used in equations such as q = mCΔT (q is heat, m is mass, C is heat capacity, ΔT is change in temperature) used to calculate the amount of heat needed to change the temperature of a substance.

    Because the equation relies on ΔT and not the absolute temperature of the system, only the relative slope of the temperature scales matters.
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