Question

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

Answer 1

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.

Answer 2

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: \(q=m*C_P*(T_f-T_i)\)