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#### apinkhighlighter

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##### Heat Transfer Question
« on: December 23, 2018, 07:26:46 AM »

In my class, we've started thermodynamics and I think I'm having trouble understanding the equation to calculate heat transfer (
q = mCpΔT ). I've supplied the problem below as well as my work. I was just wondering if someone on here would mind checking my work.

Question: How much heat is required to raise the temperature of 754 g of water from 32.9°C to 88.6°C?

q = 754(4.184J/g°C)55.7°C
q = (754)(4.184)
q = (3,154.746)(55.7)
q = 175,718.7952J

My answer is 175,718.7952 joules.

-------------

I first identified my 'm', the mass of the water at 754 grams. Then, I identified the specific heat capacity of water, which is 4.184. After I identified both the mass of the water and the heat capacity of water, I found the difference between my two temperatures, which will give me a positive 'q' value in the end. The difference between my two temperatures, 32.9°C and 88.6°C was 55.7°C. I plugged in all of my numbers into the equation ( q = mCpΔT ) and multiplied from left to right. My final answer was 175,718.7952 joules.

Thank you for reading and for your time!
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#### Borek

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##### Re: Heat Transfer Question
« Reply #1 on: December 23, 2018, 08:33:20 AM »

Looks more or less ok. You just don't have enough information to supply the answer with so many significant digits.
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#### Enthalpy

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##### Re: Heat Transfer Question
« Reply #2 on: December 24, 2018, 03:54:25 AM »

The 4184 J/kg/K apply to pure water around +20°C and 1atm. At +30°C it's rather 4180 J/kg/K and at +90°C +4205 J/kg/K
https://www.engineeringtoolbox.com/specific-heat-capacity-water-d_660.html
that's probably at the saturation pressure in that table, but at 30-90°C it makes a small difference.

Measuring a temperature (meaningfully!) to 0.1K isn't completely trivial neither, so a fourth place in the answer would be overstretched.
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