Asample of wood has a heat of combustion of 3.29 kj/g. what quantity of the wood must be burned to heat 250. g of water from 18°c to 85°c? once again, the specific heat capacity of water is 4.18 j/°c·g.

The quantity of wood that must be burned= 21.28 g

Explanation:

Heat produce by the wood = heat absorbed by the water

Q₁ = Q₂ Equation 1

Where Q₁ and Q₂ are heat produce by the wood and heat absorbed by the water respectively.

Q₁ = c₁m₁ Equation 2

Q₂ = c₂m₂(T₂-T₁) Equation 3

Substituting equation 2 and 3 into equation 1

c₁m₁ = c₂m₂(T₂-T₁) Equation 4

making m₁ the subject of the equation,

m₁ = c₂m₂(T₂-T₁)/c₁ Equation 5

Where c₁ = heat of combustion of the wood, c₂ = specific heat capacity of water, m₁ = mass of the wood, m₂ = mass of water, T₁ = initial temperature, T₂ = final Temperature.

Given: c₁ = 3.29 kJ/g = 3290 J/g, c₂ = 4.18 J/g.°C, m₂ = 250 g, T₂ = 85 °C, T₁ = 18 °C

Substituting these values into equation 4,

m₂ = 4.18×250(85-18)/3290

m₂ = 70015/3290

m₂ = 21.28 g

Thus the quantity of wood that must be burned= 21.28 g

what is the actual question though?