Question

Question 3:

(a) (6 points) We place 120.0 g of a metal at 80.00^oC in 436.3 g of water at 25.00^oC. The water is in a beaker that is also at 25.00^oC. The specific heat of water is 4.184 J K^-1 g^-1 and the heat capacity of the beaker is 0.6666 kJ K^-1. The specific heat capacity of the metal is 0.4444 J K^-1 mol^-1. What is the final temperature of the metal, the water, and the beaker?

(b) (3 points) For the decomposition of exactly one mole of a solid, X(s),

          X(s) → Y(s) + 2 Z(g)

under a constant pressure of 1.00 atm and a temperature of 25.0^oC, the value of ΔH is 58.6 kJ. What are the values of ΔU, Q, and W for this decomposition under these conditions?

Question 3: (a) (6 points) We place 120.0 g of a metal at 80.00°C in 436.3 g of water at 25.00°C. The water is in a beaker that is also at 25.00°C. The specific heat of water is 4.184 J K-1g-1 and the heat capacity of the beaker is 0.6666 kJ K-1. The specific heat capacity of the metal is 0.4444 J K-1 mol-!. What is the final temperature of the metal, the water, and the beaker? (b) (3 points) For the decomposition of exactly one mole of a solid, X(s), X(s) → Y(s) + 2 Z(9) under a constant pressure of 1.00 atm and a temperature of 25.0°C, the value of AH is 58.6 kJ. What are the values of AU, Q, and W for this decomposition under these conditions?

Answer 1

Answer a.)

Q = m*s*∆T where m= mass of substance, s = specific heat capacity and ∆T = Temperature change (Tf - Ti)

(This formula is used for metal and beaker)

Also Q=C*∆T where C= heat capacity

(This formula is used for metal)

Heat lost by metal = Heat gain by beaker + heat gain by beaker …..............(1)

Tf = is same for metal, water and beaker because zeroth law i.e.

Metal=water= beaker

Now puting given vale in equation 1

120.0g*0.4444J/(K*g)*(Tf-363) = 436.3g*4.184J/(K*g)*(Tf-298) + 0.6666*1000J/K*(Tf-298K)

53.328J/K* (Tf-363K) = 2492.0792J/K*(Tf-298)

(Tf-363K) = 46.73((Tf-298)

(Tf-363K)= 46.73*Tf - 46.73*298K

On solving

Tf = 296.63K is the final temperature of metal, water and beaker

Answer b

We know

∆H = ∆U + ∆n(gas)*RT

∆U = 58.6*1000J - 2mol*8.314J/(K*mol)*298K

∆U = 23844.856J

Or

∆U = 23.8kJ

Q at constant pressure = ∆H

So

Q = 58.6kJ

Now using first law of thermodynamics

∆U = q+ w

or

W= ∆U-q

W = (23.8-58.6)kJ = - 34.7kJ

W= -34.7kJ

Here negative sign means work is done by system which is see from reaction that producing gas is doing expansion work