The calorie (4.184 J) was originally defined as the quantity of energy required

Question

The calorie (4.184 J) was originally defined as the quantity of energy required to raise the temperature of 1.00 g of liquid water 1.00°C. The British thermal unit (Btu) is defined as the quantity of energy required to raise the temperature of 1.00 lb of liquid water 1.00°F. (a) How many joules are in 1.00 Btu? (1 lb = 453.6 g; a change of 1.0°C = 1.8°F) (b) The therm is a unit of energy consumption and is defined as 100,000 Btu. How many joules are in 1.00 therm? (c) How many moles of methane must be burned to give 1.00 therm of energy (Assume water forms as a gas): (d) If natural gas costs S0.40 per therm, what is the cost per mole of methane (Assume natural gas is pure methane)? (e) How much would it cost to warm 271 gal of water in a hot tub from 15.0°C to 42.0°C? (1 gal-3.78 L)(d of H2O = 1.00 g/mL)

Explanation / Answer

a)

Joules in BTU

1 BTU = het required to change

Q = m*C*(Tf-Ti)

1 BTU = 1 lb * (1F)

Q(joules) = 454*4.184*(1/1.8)= 1055 J

then

1 BTU = 1055 J

b)

therm = 100000 Btu = 10^5 BTU

find J

1 BTU = 1055 J

10^6BTU = x J

x = 1055*10^5 = 1.055*10^8 J

c)

methane required for this

882.0 kJ/mol

882 kJ/mol * 1 mol

mol = Q/heat combustion = (1.055*10^8)/(882*10^3)

mol of methane = 119.614

d)

0.40$ = therm

119.614mol = therm

$/mol = 0.4/119.614= 0.0033440 $ per mol

e)

cost of warming 271 gal of water

V = 271*3.78 = 1024.38 L = 1024.38kg= 1024.38*10^3 g

Q = m*C*(Tf-Ti)

Q = (1024.38*10{3)(4.184)(42-15) = 115722159.84 J

therms = 115722159.84 / ( 1.055*10^8) = 1.096 therms

0.4 $/ therm * 1.096 therm =

0.4384 $ required

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