91Ó°ÊÓ

a) Heat to be maintained inside the house for 5 days $Q=1.00×{10}^6\text{ }\frac{\text{kcal}}{\text{day}}$,

b) Temperature inside the house,$T=22°\text{C}$

c) Temperature of the water inside the barrels,$T=50°\text{C}$

d) Density of water,${\mathrm{ÒÏ}}_{water}={10}^3\frac{\text{kg}}{{\text{m}}^3}$

e) Specific heat of water,$C=1.00\text{ }\frac{\text{kcal}}{\text{kg}°\text{C}}$

The heat absorbed by the solar house is the required heat that is needed to be maintained inside the house. Thus, the amount of heat to be maintained should be constant. Now, the mass of the water considering the density given, we can get the volume of water inside the house at the maintained heat value.

Formulae:

The heat absorbed by a body due to its specific heat ( $C$), $Q=mC\mathrm{Δ}T$(i)

The volume of a body, $V=\frac{m\left(Mass\right)}{\mathrm{ÒÏ}\left(Density\right)}$ (ii)

Substitute the value of mass from equation (i) in equation (ii) with the given data, the volume of the water maintained inside the barrels can be calculated as follows:

$\begin{array}{c}V=\frac{Q}{\mathrm{ÒÏ}C\mathrm{Δ}T}\\ =\frac{(1.00×{10}^6\text{ }\frac{\text{kcal}}{\text{day}})\left(\text{5 d²¹²â²õ}\right)}{\left({10}^3\text{ }\frac{\text{kg}}{{\text{m}}^3}\right)\left(1.00\text{ }\frac{\text{kcal}}{\text{kg}°\text{C}}\right)(50°\text{C}−22°\text{C})}\\ =35.7{\text{″¾}}^3\end{array}$

Hence, the volume of water is $35.7{\text{″¾}}^3$.