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!!

Tfinal = 294.78

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

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