Specific Heat Problems Physics

specific heat problems physics
physics specific heat question?

just a little question about physics.
in a physics specific heat problem, why don’t you need to convert Kelvins into Celsius degrees?
For example, Find the amount of heat required to raise the temperature of 460 g of silver from 273 K to 300 K.
The equation is Q=mct with t being the change in temperature. The heat required is 234 J/kgC.
The book says that the change in temperature is 27 K and that’s the same as 27 Celsius degrees. I don’t understand this. Shouldn’t you change the 27 K into Celsius degrees by subtracting it from 273?

Although the Celsius and Kelvins scale differ in “where 0 is”,a degree in each scale has the same “size” . So when you only need differences in temperature, it does not matter numerically if you use Kelvin or Celsius.
In formula: K1 = C1 + 273, K2 = C2 + 273 , so
K2 – K1 = C2 + 273 – ( C1 + 273) = C2 – C1

Physics #2 – Density & Specific Heat.wmv