Question

100 g of water are cooled and the enthalpy change is -12,540 J. What is the change in temperature of the water?

Answer 1

-29.9907

Step-by-step explanation:

c = Q / (mΔT)

You're looking for ΔT, so you would plug in what you already know and get C= -12540/(100 x ΔT)

then you would solve