Specific heat of water in unit J/kg °F

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.

• 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. – David White Oct 21 '18 at 16:47
• In your first calculation, you have actually multiplied by 9 and divided by 5, even though your formula has x(5/9). – Dr Chuck Oct 21 '18 at 16:48
• But isn't °F in the denominator? I have added some more parenthesis – user600016 Oct 21 '18 at 17:05
• Yes F is in the denominator but the number you are trying to get is not. – PhysicsDave Oct 21 '18 at 17:12
• Thanks for the help everyone. I think I understodod my mistake now. – user600016 Oct 21 '18 at 17:25

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}