Solution
Concept: Wet wood acts like a sponge. The wood fiber structure remains the same volume ($V$), but water fills the pores. This increases the density without changing the bulk volume.
Let $m_{dry}$ be the mass of dry wood required and $M_{wet}$ be the mass of wet wood required.
- Heat required by the room: $Q_{room}$
- Calorific value of dry wood: $q$
From the first case (Dry Wood):
$$ Q_{room} = m_{dry} \cdot q \quad \dots (1) $$
Analyzing Wet Wood:
The wet wood has density $\rho_w$ and dry wood has density $\rho_d$.
Consider a volume $V$ of wet wood having mass $M_{wet}$.
$$ V = \frac{M_{wet}}{\rho_w} $$
The amount of actual wood fiber (dry matter) in this volume is:
$$ M_{fiber} = V \cdot \rho_d = M_{wet} \frac{\rho_d}{\rho_w} $$
The amount of water absorbed in this volume is the remaining mass:
$$ M_{water} = M_{wet} – M_{fiber} = M_{wet} \left( 1 – \frac{\rho_d}{\rho_w} \right) = M_{wet} \frac{\rho_w – \rho_d}{\rho_w} $$
Energy Balance for Wet Wood:
When wet wood is burned, the heat generated comes only from the dry fiber part. However, some of this heat is wasted to heat the absorbed water from $0^\circ$C to $100^\circ$C and then vaporize it.
$$ Q_{net} = Q_{generated} – Q_{wasted} $$
$$ Q_{generated} = M_{fiber} \cdot q $$
$$ Q_{wasted} = M_{water} [ s_{water}(100 – 0) + L_{vaporization} ] $$
Let $\Delta H = s(100) + L$.
For the room to be heated identically, $Q_{net}$ must equal $Q_{room}$ (which is $m_{dry} q$).
$$ M_{fiber} q – M_{water} \Delta H = m_{dry} q $$
Substituting expressions for $M_{fiber}$ and $M_{water}$:
$$ \left( M_{wet} \frac{\rho_d}{\rho_w} \right) q – \left( M_{wet} \frac{\rho_w – \rho_d}{\rho_w} \right) \Delta H = m_{dry} q $$
Multiply by $\rho_w$ to clear the denominator:
$$ M_{wet} [ \rho_d q – (\rho_w – \rho_d)\Delta H ] = m_{dry} q \rho_w $$
$$ M_{wet} = \frac{m_{dry} q \rho_w}{\rho_d q – (\rho_w – \rho_d)\Delta H} $$
Calculation:
Given: $m_{dry} = 34.1$ kg, $\rho_d = 600$, $\rho_w = 800$.
Water term $\Delta H = 4.2(100) + 2300 = 420 + 2300 = 2720 \text{ kJ/kg}$. (Note: $L = 2.3 \text{ MJ} = 2300 \text{ kJ}$).
Note on Calorific Value: The text states $q = 107 \text{ kJ/kg}$. This is physically too low for wood (typically 15-20 MJ/kg) and would result in a negative denominator. Based on the target answer (50 kg), $q$ is treated here as $10^4 \text{ kJ/kg}$ (10 MJ/kg), likely a typo in the print.
$$ M_{wet} = \frac{34.1 \times 10000 \times 800}{600(10000) – (200)(2720)} $$
$$ M_{wet} = \frac{272,800,000}{6,000,000 – 544,000} = \frac{272,800,000}{5,456,000} $$
$$ M_{wet} \approx 50 \text{ kg} $$