a piece of copper weighing 850 grams is placed in a cup with 450 ml of water at 21 C and the Cp of the cup is 47 J/K, how many grams of gasoline would it take to heat the entire system to 110 C?

4.2g of gasoline

Explanation:

In the problem, you need to give energy to the cup from the combustion of gasoline. The energy you need to give is:

Qcup + QWater + QCopper

As you need to increase (110ºC - 21ºC = 89º = Increase 89K) 89K, the Qcup is:

Qcup = 89K × (47J/K) = 4183J.

You can find Qwater using its specific heat, C (4.18Jg⁻¹K⁻¹), its mass (450mL = 450g) and the change of temperature, 89K:

QWater = CₓmₓΔT

QWater = 4.184Jg⁻¹K⁻¹ ₓ 450g×89K

QWater = 167569J

And Q of Copper, QCu, could be obtained in the same way (Specific heat Cu: 0.387 J/g⁻¹K⁻¹:

QCu = CₓmₓΔT

QCu = 0.387 J/g⁻¹K⁻¹ₓ850gₓ89K

QCu = 29277J

Thus, total heat you need is:

Q = Qcup + QWater + QCopper

Q = 4183J + 167569J + 29277J

Q = 201029J = 201kJ

The combustion of gasoline (Octane) produce 47.8kJ/g (Its heat of combustion). that means to produce 201kJ of energy you require:

201kJ × (1g / 47.8kJ) =