A free standing bathtub (60 x 30 inches) holds 50 gallons of water. What quantit

Question

A free standing bathtub (60 x 30 inches) holds 50 gallons of water. What quantity in grams of pure natural gas (CH4) is required to heat up the water from 70 F to 100 F?
Entalphy of Formation: CH4(g) = -74.9 O2(g) = 0 CO2(g) = -393.505 H2O = -286 A free standing bathtub (60 x 30 inches) holds 50 gallons of water. What quantity in grams of pure natural gas (CH4) is required to heat up the water from 70 F to 100 F?
Entalphy of Formation: CH4(g) = -74.9 O2(g) = 0 CO2(g) = -393.505 H2O = -286
Entalphy of Formation: CH4(g) = -74.9 O2(g) = 0 CO2(g) = -393.505 H2O = -286

Explanation / Answer

water volume in Liters = volume in gallons x 3.7851                ( where 1 gallon = 3.78541 liters)

water volume = 50 x 3.7851 = 189.255 L

water mass = voluem x density = 189.255 L x 1kg/L = 189.255 kg = 189255 g

Temp in C is TC = ( Tf-32) x ( 5/9)

Initial Temp T1    = (70-32) x ( 5/9) = 21.1 C

final Temp T2 = ( 100-32) x ( 5/9) = 37.78

now tempertaure rise or change = T2-T1 = 37.78 - 21.1 = 16.7

Heat absorbed by water = specific heat of water x mass of water x temperature rise

                     = ( 4.184 J/gK) x ( 189255 g) x ( 16.7C)

                 = 13206180 J = 13206 KJ

now the heat is given by combustion of CH4 which is given by equation

CH4 + 2O2 ---> CO2 + 2H2O

dH reaction = dH Products - dH reactnats

               = dH CO2 + 2dH H2O - dH CH4 - dH O2

            = -393.5 + 2(-286) -(-74.9) - 0

               = 890.6 KJ/mol

Number fo moles of CH4 required = total heat / Heat of combustion of CH4 per moles

                                          = 13206 /890.6    = 14.828

         CH4 mass = moles x molar mass = 14.828 x 16.04 g/mol   = 238 g

Thus 238 grams of CH4 needed to provided required heat

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