Example 8.3 sixth edition

[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?

[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)

[Jason Ye 2I]

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

[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).