Question

How many calories are required to heat 62.0 kilograms ice from 0 degrees celsius to 47.0 degrees celsius?

2 C₈H₁₈ + 25 O₂ --> 16 CO₂ + 18 H₂O
How many grams of CO₂ is released inot the atmosphere by the complete combustion of 40.0 gallons of gasoline? Assume the formulat for gasoline is C₈H₁₈ with a density of 0.703 g/mL.

Answer 1

This problem needs to be broken up into a couple of steps as -

1. heat required to melt 62 Kgs. or 62000 gms . of ice at 0^o C

Q = m L

.... = 62000 x 80

.... = 4960000 calories

[ L is latent heat of ice = 80 cal / gm , Q represents heat required, m is mass of water in gms =62000gms.]

2. Heat required to raise the temperature of water from 0^o C to 47^o C

Q^'  = mc( t_final - t_initial )

.... = 62000 x 1 x (47 - 0 )

..... = 2914000 calories

Thus, total heat required to heat 62 Kilograms of ice from 0^o C to 47^o C

....................................................................... = ( Q + Q^' ) cal.

....................................................................... =( 4960000 + 2914000 ) cal.

..........................................................................= 7874000 calories

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Glad to help. Please submit the second question separately as fresh question for detailed answer.

However, the following hint can be useful-

Use stoichiometric relation to get the mass of CO2 released from given equation.

the answer to second question is,-

grams of CO2 released in atmosphere = 328674.68 gms