If you used the 2600 you expend in energy in one day to heat 7.0×104 of water at 20, what would be its new temperature?

Answer 1

I'm going to do some assuming here.

2600= 2,600 cal

7.0×10^4=7.0×10^4 g

and 20=20ºC

Use the following formula:

q=mc∆T

Where

q=2,600 cal
m=7.0×10^4 g
c= 1.00 cal/g °C
∆T=Tf-Ti=Tf-20ºC

solve for Tf

q/mc=Tf-20ºC

{2,600 cal/[(7.0×10^4 g)*(1.00 cal/g°C)]}+20ºC=Tf

*****if 2,600 is kcal, just change 2.6 x 10^6 to get the correct answer.

Answer 2

Note should say if 2,600 is kcal, just change it to 2.6 x 10^6 to get the correct answer.