## Sunday, October 17, 2010

1.                  Phileas Fogg, the character who went around the world in 80 days, was very fussy about his bathwater temperature. It had to be exactly 38.0o C. You are his butler, and one morning while checking his bath temperature, you notice that it’s 42.0oC. You plan to cool the 100.0 kg of water to the desired temperature by adding an aluminum-duckie originally at freezer temperature (-24.0oC). Of what mass should the Al-duckie be? [Specific heat of Al = 0.900 J/(goC); density of water =1 .00 g/ml]. Assume that no heat is lost to the air.

## Solution #1

-Qwater QAl
mwcwDTw = - mAlcAlDTAl
100 000g(4.19 J/[g C] (38.0 – 42.0)= - mAl (0.900 J/[g C])(38.0 – [-24.0]).
mAl = 30036 g = 30.0 kg.

2.                  A certain material’s(environment) temperature increases by 1.0oC for every 1560 J that it gains. A 0.1964 g sample of quinone (molar mass = 108.1 g/mole) was burnt, and the surrounding material’s temperature increased from 20.3 oC to 23.5 oC. Find the molar heat of combustion for quinone.

## Solution #2

Qmaterial = 1560 J/oC (23.5-20.3) oC  = 49 92 J = 5.0 kJ
Note that 1560 J/oC is equivalent to mc from Q = mcDT.
DH = -Q = -5.0 kJ
n = 0.1964g/[108.1 g/mole] 0.00182 mole
DH/n = -5.0 kJ/0.00182 mole = -2.7 X 103 kJ/mole of quinine

3.                  A 1.55 g of CH4O sample is burnt in a calorimeter. If the molar heat of combustion of CH4O is -725 kJ/mole, and assuming that the 2.0 L of water absorbed all of the heat of combustion, what temperature change did the water experience?

## Solution #3

n = 1.55g/[32g/mole] = 0.0484 moles
n[DH/n] = DH
0.0484 moles(-725 kJ/mole)= - 35.1 kJ
Q = - DH
Q = 35.1 kJ = 35 100 J
Q = mc DT
35 100 = 2000g (4.19 J/[g C]) DT
DT = 4.2 C

4.                  0.20 moles of HX were neutralized by NaOH. The concentrations of the base and acid were equal. If the temperature of the water in the calorimeter increased from 19.9 to 24.6 C, what was the original concentration of HX?
Molar heat of neutralization = -80 kJ/mole of HX

## Solution #4

n[DH/n] = DH
0.20mole [-80 kJ/mole] = -16 kJ.
Q = - DH
Q = 16 kJ = 16000 J.
Q = mc DT
16000 = m (4.19 J/[g C])( 24.6 – 19.9)
m = 812g of water.
Since the concentrations of the acid and the number of moles were equal (moles are equal because of the 1:1 ratio in which HX reacts with NaOH), then the volumes used were equal.
812 g = 812 mL created from 812/2 = 411 mL of acid = 0.411 L
concentration = n/V = 0.20 moles/0.411 L = 0.49 M.

5.                  In real calorimeters, most of the heat released by the bomb is absorbed by water, but a certain amount is also absorbed by the metal and insulation surrounding the water tank. A certain calorimeter absorbs 24 J/oC. If 50.0 g of 52.7oC water is mixed with the calorimeter’s original 50.0 g of 22.3oC water, what will be the final temperature of the mixture?

## Solution #5

The heat lost by the hot water will be gained by the cold water and by the calorimeter.
-Qhot Qcold Qcalor
-mc DT = mc DT + 24J/0C DT
-50(4.19)( x – 52.7) = 50(4.19)(x –22.3) + 24(x – 22.3)
-209.5( x – 52.7) = 233.5(x –22.3)
-0.897( x – 52.7) = x – 22.3
-0.897 x  + 47.27 = x – 22.3
69.57 = 1.897 x
x = 37 oC.

## Saturday, October 16, 2010

### SPECIFIC HEAT PROBLEMS

1. What is the specific heat of a substance that absorbs 2.5 x 103 joules of heat when a sample of 1.0 x 104 g of the substance increases in temperature from 10.0C to 70.0C?

q = (m)(Dt)(cp)   where q  is the energy in joules
m is the mass

Dt is the change in initial temperature to the final temperature
cp is the specific heat (heat required to raise 1 g of the substance
1C)

q = (m)(Dt)(cp)
2.5 x 103 J = (1.0 x 104 g)(70C-10C)(cp)

[2.5 x 103 J]                 = cp
[1.0 x 104 g][70C-10C]
cp=   0.0042 J
gC

2) A 1.0 kg sample of metal with a specific heat of 0.50 KJ/KgC is heated to 100.0C and then placed in a  50.0 g sample of water at 20.0C.  What is the final temperature of the metal and the water?
qmetal = (m)(Dt)(cp)      qwater = (m)(Dt)(cp)
qmetal  =  qwater
(m)(Dt)(cp)    =     (m)(Dt)(cp)
[1.0Kg][100C-Tf][0.50KJ]  =  [50.0g][Tf-20C][4.184 J]
[   KgC]                                 [      gC ]
[1.0Kg][1000g][100C-Tf][0.50KJ][   1Kg][1000J]  =  [50.0g][Tf-20C][4.184 J]
[   1Kg]                 [   KgC][1000g][   1KJ]                                [      gC ]
[1000g][100C-Tf][0.50J]  =  [50.0g][Tf-20C][4.184 J]
[   gC]                                [      gC ]
[1000g][100C-Tf][0.50   J/gC]  = [Tf-20C]
[ 50.0g]                [4.184 J/gC]
[2.39][100C-Tf]  =  [Tf-20C]
239C-2.39Tf  =  Tf-20C
239C + 20C  =   T+  2.39Tf
259C  =  3.39Tf
259C  =  Tf

3.39
76C  =  Tf
Answer: Final temperature of metal and water is 76C

3)  3) A 2.8 kg sample of a metal with a specific heat of 0.43KJ/KgC is heated to 100.0C then placed in a 50.0 g sample of water at 30.0C.  What is the final temperature of the metal and the water?

qmetal = (m)(Dt)(cp)      qwater = (m)(Dt)(cp)
qmetal  =  qwater
(m)(Dt)(cp)    =     (m)(Dt)(cp)
[2.80Kg][100C-Tf][0.43KJ]  =  [50.0g][Tf-30C][4.184 J]
[   KgC]                                 [      gC ]
[2.8.0Kg][1000g][100C-Tf][0.43KJ][   1Kg][1000J]  =  [50.0g][Tf-30C][4.184 J]
[   1Kg]                 [   KgC][1000g][   1KJ]                                [      gC ]
[2800g][100C-Tf][0.43J]  =  [50.0g][Tf-30C][4.184 J]
[   gC]                                [      gC ]
[2800g][100C-Tf][0.43   J/gC]  = [Tf-30C]
[ 50.0g]                [4.184 J/gC]
[24.08][100C-Tf]  =  [Tf-30C]
2408C-24.08Tf  =  Tf-30C
2408C + 30C  =   T+  24.08Tf
2438C  =  25.08Tf
2438C  =  Tf

325.08
97C  =  Tf
Answer: final temperature of metal and water is 97C