Question

Please help if you can

If 980 kJ of enegy are added to 6.2 L of water at 291 K, what will the final temperature of the water be?
Thank you!!

Answer 1

Tfinal = 294.78

Step-by-step explanation:

We will use the following equation, with q=heat, m=mass, c=specific heat, ΔT= change in temperature:

q = mcΔT

q = 980 kJ = 98000 J

m = 6200 g (convert from liters)

c = 4.186 J/g (this is a standard to memorize)

ΔT = Tfinal - Tinitial = Tfinal - 291 K

98000 = 6200*4.186*(Tfinal - 291)

3.776 = Tfinal - 291

Tfinal = 294.78