Example 8.3 sixth edition

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MichelleRamirez_2F
Posts: 63
Joined: Fri Sep 28, 2018 12:28 am

Example 8.3 sixth edition

Postby MichelleRamirez_2F » Tue Feb 12, 2019 2:34 pm

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?

705192887
Posts: 77
Joined: Fri Sep 28, 2018 12:18 am

Re: Example 8.3 sixth edition

Postby 705192887 » Tue Feb 12, 2019 2:44 pm

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)

Jason Ye 2I
Posts: 33
Joined: Fri Sep 28, 2018 12:22 am

Re: Example 8.3 sixth edition

Postby Jason Ye 2I » Tue Feb 12, 2019 2:46 pm

The changes in kelvin are the same as the changes in celsius.

Chloe Likwong 2K
Posts: 70
Joined: Fri Sep 28, 2018 12:23 am

Re: Example 8.3 sixth edition

Postby Chloe Likwong 2K » Tue Feb 12, 2019 2:49 pm

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).


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