What is the total number of joules of heat energy absorbed by 25 grams of water when it is heated from 50 °C to 70 °C? Specific heat of water is 4.18/J/g °C and q=Mc △T

Question

anonymous · Answer

I think you might have typed the formula incorrectly. I am used to using $$q= m \times \Delta T \times Cp$$
the q is the  heat in J
Cp is the specific heat
Delta T is the change in temp in K
and m is the mass in g
Plug in all the numbers accordingly and you should get your answer.

aaronq · Answer

The order of m, $\Delta T$ and $C_P$ doesn't really matter since they're multiplied, and the temperature doesnt need to be in Kelvin since were only using the $\sf difference$ in temperature (and the scale of Kelvin is equally spaced with celsius). But i'm glad you explicitly identified the variables, @noelleheart  

you should also note that $\Delta T= T_f-T_i$

so the formula is:   $q=m*C_P*(T_f-T_i)$