Question

An iron boiler of mass 180 kg contains 830 kg of water at 11 ∘C. A heater supplies energy at the rate of 58,000 kJ/h. The specific heat of iron is 450 J/kg⋅C∘, the specific heat of water is 4186 J/kg⋅C∘, the heat of vaporization of water is 2260 kJ/kg⋅C∘. Assume that before the water reaches the boiling point, all the heat energy goes into raising the temperature of the iron or the steam, and none goes to the vaporization of water. After the water starts to boil, all the heat energy goes into boiling the water, and none goes to raising the temperature of the iron or the steam.

Part A How long does it take for the water to reach the boiling point from 11 ∘C? Express your answer using two significant figures.

Part B How long does it take for the water to all have changed to steam from 11 ∘C? Express your answer using two significant figures.

Answer 1

Mass of iron - m_i = 180 kg

Mass of water = m_w = 830 kg

Initial temperature of iron and water = T₁ = 11 ^oC

Boiling temperature of water = T₂ = 100 ^oC

Specific heat of iron = C_i = 450 J/(kg.^oC) = 0.45 kJ/(kg.^oC)

Specific heat of water = C_w = 4186 J/(kg.^oC) = 4.186 kJ/(kg.^oC)

Heat of vaporization of water = L = 2260 kJ/kg

Energy supplied by the heater = P = 58000 kJ/h

Time taken by water to reach boiling point from 11^oC = t₁

Time taken by water to completely change to steam from 11^oC = t₂

Pt₁ = m_iC_i(T₂ - T₁) + m_wC_w(T₂ - T₁)

58000(t₁) = (180)(0.45)(100 - 11) + (830)(4.186)(100 - 11)

t₁ = 5.46 hours

Pt₂ = m_iC_i(T₂ - T₁) + m_wC_w(T₂ - T₁) + m_wL

58000(t₂) = (180)(0.45)(100 - 11) + (830)(4.186)(100 - 11) + (830)(2260)

t₂ = 37.80 hours

A) Time taken by the water to reach boiling point from 11^oC = 5.46 hours

B) Time taken by water to all have changed to steam from 11^oC = 37.80 hours