Let's break down and solve the problem step by step: --- **Given Data:** - Thermal conductivity, \( k = 4.8 \times 10^{-4} \) kW/m·°C - Wall thickness, \( d = 20.5 \) cm \( = 0.205 \) m - Inside temperature, \( T_{in} = 24.0^\circ C \) - Outside temperature, \( T_{out} = 0.0^\circ C \) - Area, \( A \): *NOT provided*, so let's denote it as \( A \) (the answer will be in terms of \( A \)) - Heat of combustion of gas, \( Q_{comb} = 9300 \) kcal/m³ - Time, \( t = 1 \) day \( = 24 \times 3600 = 86400 \) s --- ### **Step 1: Calculate the Heat Loss Per Second (Power Required)** The rate of heat loss through the walls and roof (Fourier's law): \[ \dot{Q} = \frac{kA(T_{in} - T_{out})}{d} \] Plug in the values: \[ \dot{Q} = \frac{(4.8 \times 10^{-4})A \times (24.0 - 0.0)}{0.205} \] \[ \dot{Q} = \frac{(4.8 \times 10^{-4})A \times 24.0}{0.205} \] \[ \dot{Q} = \frac{0.01152A}{0.205} \] \[ \dot{Q} \approx 0.0562A \text{ kW} \] --- ### **Step 2: Calculate Total Heat Lost in One Day** \[ Q_{total} = \dot{Q} \times t \] \[ Q_{total} = (0.0562A) \times 86400 \text{ seconds} \] Since 1 kW = 1 kJ/s, and 1 kcal = 4.184 kJ, \[ Q_{total} = 0.0562A \times 86400 = 4854.7A \text{ kJ} \] Convert kJ to kcal: \[ Q_{total} (\text{kcal}) = \frac{4854.7A}{4.184} \approx 1160A \text{ kcal} \] --- ### **Step 3: Calculate Volume of Gas Required** Each m³ of gas gives 9300 kcal. \[ \text{Volume of gas needed} = \frac{Q_{total} (\text{kcal})}{9300} \] \[ \text{Volume} = \frac{1160A}{9300} \] \[ \text{Volume} \approx 0.125A \text{ m}^3 \] --- ## **Final Answer** **If the total surface area \(A\) of the walls and roof is known, the number of cubic meters of gas that must be burned each day is:** \[ \boxed{0.125A} \] where \(A\) is in square meters. --- ### **If you provide the area \(A\), I can give a numerical answer!** --- #### **Summary of Steps:** 1. Used Fourier’s law to find heat loss rate. 2. Multiplied by total seconds in a day. 3. Converted energy to kcal. 4. Divided by energy released per m³ of gas. --- **If you have the value of the area \(A\), plug it into the final formula to get the volume of gas needed per day.**