Question

Ethylene glycol has a specific heat of 0.578 cal/(g×°C). If 23.2 g of ethylene glycol absorbs 75.6 cal of heat energy, what will the temperature increase be?

Answer 1

q = mass x specific heat x (Tfinal-Tinitial)
75.6 = 23.3 x 0.578 x delta T.
Solve for delta T.