A 25-kg cast-iron wood-burning stove contains 5kg of soft pine wood and 1 kg of air. All the masses are at room temperature, 20 degree C, and pressure, 101 kPa. The wood now burns and heats all the mass uniformly, releasing 1500W. Neglect any air flow, changes in mass of wood, and heat losses. Find the rate of change of the temperature (dT/dt) and estimate the time it will take to reach a temperature of 75 degree C. Assume the iron, wood, as well as air to be incompressible substances (note that this is not a correct assumption, as we will see later in the semester, but it is useful for applying the concepts related to the First Law of Thermodynamics). c_iron = 0.42 kJ / (kg. degree C) : c_wood = 1.38 kJ / (kg. degree C) : c_air = 0.717 kJ / (kg. degree C)

dT / dt = 0.0828 K / s

Time taken (dt) = 664.29 s ... 11.07 min

Explanation:

Assuming all the energy lost by burning interchanges within the system and No heat loss from the system.

Data Given:

Q = 1500 W

m_air = 1 kg

m_iron = 25 kg

m_wood = 5 kg

Cv_air = 0.717 KJ/k C

Cv_iron = 0.42 KJ/k C

Cv_wood = 1.38 KJ/k C

part a

Using Energy Balance:

m_air*Cv_air * (dT/dt) + m _iron*Cv_iron * (dT/dt) + m_wood*Cv_wood * (dT/dt) = Q / 1000

dT / dt = (Q / 1000) / (m_air*Cv_air + m _iron*Cv_iron + m_wood*Cv_wood)

Plug the given data in equation above and solve for dT/dt:

dT / dt = (1500 / 1000) / (1*0.717 + 25*0.42 + 5*1.38)

dT / dt = 0.0828 K / s

Answer: dT / dt = 0.0828 K / s

part b

dT = (75 - 20) = 55 K

Using the derived expression for dT / dt in part a:

55 / dt = 0.0828

dt = 55 / 0.0828 = 664.29 s ... 1.1.07 min

Answer: dt = 664.29 s ... 11.07 min to reach T = 75 C