How many kilograms of oil are needed to heat 190 kg of water from 22 ∘C to 100∘C? (Assume that 1lb=454g, the specific heat of liquid water is 4.184 J/g∘C.)

m = 1.3 Kg

Explanation:

according to Specific Heat Formula:

Q = mc∆T

when Q is the heat energy in Joules

m is the mass of water

C is the specific heat its unit J/kg∙K

∆T change in temperature (the final t - the initial t)

by substitution:

Q = 190000 g * 4.184 J/g∘C * 78

Q = 62006880 J

when The combustion of 1.0 lb of oil provides 2.2*107J

So lb of oil when Q = 62006880 J = 2.8 Ib

and when 1lb=454g

So 2.8 Ib = 1271.2 g = 1.3 Kg