# Chapter 3 - Matter and Energy - Exercises - Cumulative Problems - Page 91: 101

$6.05 kWh$

#### Work Step by Step

Using the equation $q=mcΔT$, we can solve for the heat. The specific heat capacity of water is $4.184\frac{J}{g\times^{\circ}C}$and the mass of the water in this case is 55 gallons which equals 208.198 L because $1 gallon = 3.78541 L$ so we can do $55 gallons\times 3.78541 L = 208.198 L$. Therefore we have 208.198 kg or 208198 grams since we are told we have 208.198 L, and since the density of water is 1 g/L we can use the simple relationship to get 208198 grams. $q = (208198 grams\times4.184\frac{J}{g\times^{\circ}C}\times25) = 21777511 J$ Now lastly we must convert the heat from Joules to kWh using the following relationship. $1kWh=3.6\times10^{6}J$ $21777511 J\div 3.6\times10^{6}J =6.05 kWh$

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