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We know that specific heat of water is 4186 J/kg °C This is the amount of heat per unit mass required to raise/change the temperature by one degree Celsius.

We also know that ∆°C = 5/9 °∆F = ∆K (comparing change of temperature 1°C with 1°F and 1K).

So will specific heat of water also be 4186J/ (kg *(5/9)°F) = 7524J/ (kg °F)?

Thinking about it logically, celcius scale as 100 divisions whereas Farenheit scale has 180 divisions.Assuming length of both scales are same, we would require less heat to raise by 1 division in Farenheit scale due to more divisions so smaller length of each division. By ratio and proportion, we get s = 4186 * 5/9 J/(kg °F) = 2322.22 J/(kg °F).

Which is correct? I couldn't find the value in this scale anywhere.

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    $\begingroup$ You want to multiply the 4186 value by 5/9, but why in the world would you want "mixed" units? The value needs to be either all English units or all metric units, but not both. $\endgroup$ – David White Oct 21 '18 at 16:47
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    $\begingroup$ In your first calculation, you have actually multiplied by 9 and divided by 5, even though your formula has x(5/9). $\endgroup$ – Dr Chuck Oct 21 '18 at 16:48
  • $\begingroup$ But isn't °F in the denominator? I have added some more parenthesis $\endgroup$ – user600016 Oct 21 '18 at 17:05
  • $\begingroup$ Yes F is in the denominator but the number you are trying to get is not. $\endgroup$ – PhysicsDave Oct 21 '18 at 17:12
  • $\begingroup$ Thanks for the help everyone. I think I understodod my mistake now. $\endgroup$ – user600016 Oct 21 '18 at 17:25
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Everyone gets angry at unit conversion problems sometimes. When this happens to you, grab a napkin where you can write proper-looking fractions, and carry the units around with their corresponding values. If the algebra doesn't cancel out the units you want to remove, then your unit conversion is not correct.

If you have fraction like $5/9$ in the denominator of another fraction, the odds that you get it in the calculator correctly are 50-50. If you have a PhD and you write a fraction in the denominator, the odds that you get it in the calculator correctly are less than 50-50.

\begin{align} 4186 \frac{\rm J}{\rm kg\,^\circ C} &= 4186 \frac{\rm J}{\rm kg\,^\circ C} \color{gray}{\times \frac{100\,^\circ \rm C}{180\,^\circ\rm F}} \\&= 2326 \frac{\rm J}{\rm kg\,^\circ F} \end{align}

Your "thinking about it logically" paragraph contains correct reasoning.

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  • $\begingroup$ Oh I understood my mistake now. Thank you very much :) $\endgroup$ – user600016 Oct 21 '18 at 17:24

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