Question

1/ Determine the amount of heaat required to raise the temperature of 2500 grams of water form 35C degree to 57C degree. The specific heat of water is 4.18J/g*C.

2/A system releases 350. kilojoules of heat while 750. kilojoules of work are done on it. Determine the change in internal energy (△ E) of the system

3/ Afixed mass of gas is placed in a cylinder fitted with a movable piston. 2000.joules of heat are added to the gas, causing it to expand. The expanding gas caused the piston to rise, thus increasing the volume of the cylinder 4,5 liters. If the constant downward pressure of the piston on the gas was 3.5 atmospheres.Calculate the (△ E) of the gas.

4/ How many joules of heat are released to the surroundings when145.0 grams of aluminum react with an excess amount of iron(III) oxide according to the following reaction?

2AL + Fe2O3 ------> AL2O3 + 2Fe △Hrxn= -849KJ

5/ use the following reactions 1,2 and 3 to determine the enthalpy of reaction for:

N2O(g) + NO2(g) -----> 3NO(g) △Hrxn= ?

I/ N2(g) + O2 (g) ------> 2NO(g) △Hrxn=+180.7KJ

II/ 2NO(g) + O2(g) ------> 2NO2(g) △Hrxn=-113.1KJ

III/ 2N2O(g) ------> 2N2(g) + O2(g) △Hrxn=-163.2KJ

6/ A 75 gram piece of copper at 100C degree is placed in an insulated container that contains 225 grams of water at 35C degree. Determine the equilibrium temperature reached in the copper water system. The specific heat of copper is 0.385J/g*C and the specific heat of water is 4.18J/g*C. You may assume that there is no exchange of heat between the system and surroundings.

Answer 1

1) heat,q=mcT=2500 g*4.18J/g*C* (57-35)C=229900J

2)U=q + W    W=work done

         =350 kj +750 kj=1100kj

3) U=q-pV+W(non-expansion)=2000J-3.5 atm* 101325 Nm2/atm *1000m3=2000J-354.64J=1645.3J

V=5-4=1 dm3=1 (10-1)3 m3=10^-3m3

(1 atm = 101325 Pa=N/m2)

4)molar mass of Al=27g/mole

2 moles of Al=2*27g=54 g Al reacts with excess amount of iron(III) oxide to give heat=-849KJ

                               so,145 g Al reacts to give heat=-849kj/54g *145g=-2279.7kj

5) 2N2O(g) ------> 2N2(g) + O2(g)

N2O(g) ------> N2(g) + 1/2 O2(g) △Hrxn=-163.2KJ/2=-81.6kj.....(.1)

2NO(g) + O2(g) ------> 2NO2(g)

NO2(g)-----------NO(g) + 1/2 O2(g) △Hrxn=113.1KJ/2=56.55kj.......(2)

N2(g) + O2 (g) ------> 2NO(g) △Hrxn=+180.7KJ.........(3)

now eqn 1+2+3gives   N2O(g) + NO2(g) -----> 3NO(g) △Hrxn= -81.6kj+56.55kj+180.7KJ.=155.65Kj

6)q=mcT

75 g * 0.385J/g*C*(100-T)C=225 g*4.18J/g*C*(T-35)(heating)

T= equiibrium temp

100-T/T-35=32.57

100-T=32.57T-1139.95

1239.95=33.57T

T=36.9C