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Example 8.3 sixth edition

Posted: Tue Feb 12, 2019 2:34 pm
by MichelleRamirez_2F
In the sixth edition textbook the example problem for specific heat and molar heat states that the temperature changes from 20C to 100C. In their work however they treated the deltaT as 80.K. My question is why they didn't convert 80C to 353K?

Re: Example 8.3 sixth edition

Posted: Tue Feb 12, 2019 2:44 pm
by 705192887
They don't need to convert it because Celsius is Just kelvin—273. Therefore an increase of 80°C is also an increase of 80°K.
∆T(°C) = ∆T(K)

Re: Example 8.3 sixth edition

Posted: Tue Feb 12, 2019 2:46 pm
by Jason Ye 2I
The changes in kelvin are the same as the changes in celsius.

Re: Example 8.3 sixth edition

Posted: Tue Feb 12, 2019 2:49 pm
by Chloe Likwong 2K
In the problem, it doesn't matter whether you convert it or not because you're simply finding the difference (whether you add 273 or not, the difference between the two temperatures would still be 80).