Ignoring any losses, estimate how much energy (in units of btu) is required to raise the temperature of water in a 90-gallon hot-water tank from 60°f to 110°f. the specific heat of water is approximated as a constant, whose value is 0.999 btu/·lbmr at the average temperature of (60 + 110)/2 = 85ºf. in fact, c remains constant at 0.999 btu/lbm·r (to three digits) from 60ºf to 110ºf. for this same temperature range, the density varies from 62.36 lbm/ft3 at 60ºf to 61.86 lbm/ft3 at 110ºf. we approximate the density as remaining constant, whose value is 62.17 lbm/ft3 at the average temperature of 85ºf.

Q=36444.11 Btu

Explanation:

Given that

Initial temperature = 60° F

Final temperature = 110° F

Specific heat of water = 0.999 Btu/lbm.R

Volume of water = 90 gallon

Mass = Volume x density

Mass ,m= 90 x 0.13 x 62.36 lbm

m=729.62 lbm

We know that sensible heat given as

Q= m Cp ΔT

Now by putting the values

Q= 729.62 x 0.999 x (110-60) Btu

Q=36444.11 Btu

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hope this is what you're talking about