Answer to Question #312781 in General Chemistry for Krix

Question

First Law of Thermodynamics and Calorimetry (5 points each)

A gas system with constant pressure has change in internal energy of 440 J. Heat was lost to the surroundings and 250 J work was done to the surroundings as well. How many calories of heat was lost?

0.315 moles of butane (C4H10) was burned in a copper calorimeter containing 4500 g of water. The calorimeter weighs 2,500 g. The temperature was observed to increased by 3.15 °C. Calculate the molar heat of combustion of butane (HEAT OF COMBUSTION PER MOLE). The specific heat of copper and water are 0.386 J/g. °C and 4.18 J/g. °C respectively.

A 45-g silver spoon (specific heat 0.233 J/g °C) at 25 °C is placed in 180 g of coffee at 85 °C and the temperature of the two become equal. What is the final temperature when the two become equal? The coffee has specific heat of 5.12J/g. °C.

Answer 1

$\Delta$ U = Q - W;

Q = $\Delta$ U + W = 440 + 250 = 690 J = 164.9 cal.

$\Delta$ H = -m * c * $\Delta$ T;

m = 4500 g;

c = 4.18 J/(g * °C);

$\Delta$ T = 3.15 °C;

$\Delta$ H = -4500 * 4.18 * 3.15 = 59521.5 J/mol = 59.5 kJ/mol;

For n(C₄H₁₀) = 0.315 moles:

$\Delta$ H' = $\Delta$ H * 0.315 = 59.5*0.315 = 18.7 kJ/mol.

Q_Al = - Q_coffee;

m_Al * c_Al * (T_final - T_Al) = m_coffee * c_coffee * (T_final - T_coffee);

T_final = (m_Al * c_Al * T_Al) + (m_coffee * c_co*ffee * T_coffee) / ((m_Al * c_Al) + (m_coffee * c_coffee)) = (45 * 0.233 * 25) + (180 * 5.12 * 85) / ((45 * 0.233) + (180 * 5.12)) = 262.125 + 78336 / (10.485 + 921.6) = 78598.125/932.085 = 84.3 °C.

Answer:

• 164.9 cal;
• 18.7 kJ/mol;
• 84.3 °C.