Question 1:
85.0 mL of H2O is initially at room temperature (22C). A chilled steel rod at 2C is placed in the water. If the final temperature of the system is 21.2C, what is the mass of the steel bar?
Specific heat of water = 4.18 Specific heat of steel = 0.452
Answer:
THis is a conservation of energy problem; to solve add up thermal energies before and after.
85 x 4.184 x 22 + MS x .452 x 2 = 85 x 4.184 x 21.2 + MS x .452 x 21.2
MS = mass of steel
Solve for MS.
Or you could say that the energy gained by the steel = the energy given up by the water
MS x .452 x [21.2 -2] = 85 x 4.184 x [22 - 21.2]
Question 2
2 HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2 H2O(l) ΔH = −118 kJ
Calculate the heat evolved when 96.4 mL of 0.500 M HCl is mixed with 300.0 mL of 0.445 M Ba(OH)2.
1 kJ
Assuming that the temperature of both solutions was initially 25.0°C and that the final mixture has a mass of 396.4 g and a specific heat capacity of 4.18 J °C-1 g-1, calculate the final temperature of the mixture.
Answer :
96.4 mL of 0.500 M HCl = 0.0482 moles HCl
300.0 mL of 0.445 M Ba(OH)2. = 0.1335 moles Ba(OH)2
by the reaction
2 HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2 H2O(l)
they needed twice as many moles of HCl as there were moles of Ba(OH)2
they did not add enough HCl
HCl is the limiting reagent
find heat:
0.0482 moles HCl @ 118 kJ/2 moles HCl = 2.844 kiloJoules
find dT
dH = mCdT
2,844 joules = 396.4g (4.18 J/g-C)dT
dT = 1.72 hotter
the final temp is 26.7Celsius
Question 3 :
A 0.040 kg ice cube at 0 degrees C is placed in an insulate box that contains 0.0075 kg of steam at 100 degrees C. When the thermal equilibrium is established, all of the ice has melted. Find the final temperature of the closed system. The specific heat of water is 4186 J/kg*K. The latent heat of fusion for ice is 3.34X10^5 J/kg. The latent heat of vaporization for water is 22.6X10^5 J/kg.
Choices:
*22.7 C
*33.6 C
*44.9 C
*50.7 C
*66.4 C
Answer :
Heat given off by steam = Heat absorbed by ice
Qs = Qi
Qs = 0.0075(22.6 x 10^5) + 0.0075(4186)(100 - T)
Qi = 0.040(3.34 x 10^5) + 0.040(4186)(T - 0)
where
T = equilibrium temperature of the system
Since Qs = Qi,
0.0075(22.6 x 10^5) + 0.0075(4186)(100 - T) = 0.040(3.34 x 10^5) + 0.040(4186)(T - 0)
0.1695 x 10^5 + 31.395(100 - T) = 0.1336 x 10^5 + 167.44(T)
Simplifying,
0.1695 x 10^5 + 3139.5 - 31.395T = 0.1336 x 10^5 + 167.44T
Combining terms,
167.44T + 31.395T = 0.1695 x 10^5 - 0.1336 x 10^5 + 0.031395 x 10^5
198.835T = 6729.5
and solving for T,
T = 33.8 C
Question 4:
A 19.0 g sample of ice at -17.0°C is mixed with 102.0 g of water at 86.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J/g°C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
Answer :
19.0 g ice at -17.0 deg going to 0 deg requires-
19.0g x 2.08 J/g deg x 17 deg = 671.8 J
Melting 19.0 g ice at 0 deg requires-
19.0g x (1/18.02 g/mol) x 6.02kJ/mol x 1000 J/kJ x = 6347.4 J
671.8 J + 6347.4 J = 7019.2 J
That energy will come from cooling 102.0 g H20-
7019.2 J = 102.0g x 4.18 J/g deg x (86.0 - t)
t = 69.5 deg
At this point you have 102.0 g water at 69.5 and 19 g water at 0 deg
102.0 g x 4.18 x (69.5 - t) = 19.0 x 4.18 x (t - 0)
t = final temperature = 58.6 deg C
Question 5:
A 16.7 g sample of ethanol at 230.5 K is mixed with 55.6 g of ethanol at 317.1 K. Calculate the final temperature of the mixture assuming no heat is lost to the containers and surroundings. The specific heat of ethanol is 2.44J/K·g.
Answer :
The temperature of the first solution will increase and the temperature of the second solution will decrease
16.7 g x 2.44 ( x - 230.5 ) = 55.6 g x 2.44 ( 317.1 -x)
40.75 ( x - 230.5 ) = 135.7 ( 317.1 - x )
40.75 x - 9392.9 = 43030.5 - 135.7 x
176.4 x = 52423
x = 297 K
Question 6:
Calculate the amount of heat required to completely sublime 25.0 g of solid dry ice (CO2 ) at its sublimation temperature. The heat of sublimation for carbon dioxide is 32.3 kJ/mol.
Answer :
First, determine how many moles of CO2 there are. The mol. wt. of CO2 is 12 + 2 x 16 = 44 g/mol.
So, 25.0 g equals 25/44 = 0.568 moles.
So the heat required is 0.568 mol x 32.3 kJ/mol = 18.3 kJ
Question 7:
A 188 g sample of a substance is heated to 323°C and then plunged into a 100 g aluminum calorimeter cup containing 166 g of water and a 18 g glass thermometer at 11.5°C. The final temperature is 35.3°C. What is the specific heat of the substance?
Answer :
Heat out Qiut from the substance = Q into the cup, thermometer & water
So Q out = m*c*deltaT = 0.188kg*c*(323-35.3) = 54.1c
Q in = m*c*deltaT for each substance = 0.100kg*900J/kg-oC*(35.3-11.5) + 0.166kg*4200*(35.3-11.5) + 0.018*840*(35.3-11.5) = 19165J
So 54.1c = 19165 => c = 19165/54.1 = 354J/kg-oC
Question 8:
A 448 g block of metal absorbs 5586 J of heat when its temperature changes from 19*C to 42*C.
Calculate the specific heat of the metal. Answer in units of J/kg. *C.
Answer :
delta Q = mc delta T
5586 = 0.448 x 23 x c
c = 542 J/kg.*C
Question 9:
A 500.0 g sample of an element at 195 degrees C is dropped into an ice-water mixture; 109.5 g of ice melts and an ice-water mixture remains. Calculate the specific heat of the element.
Answer :
The latent heat of fusion of ice is 334kJ/kg
Thus, Q=0.1095 kg x 334 kJ/kg = 36.573 kJ
C= 36.573 kJ / 0.5kg / 195K
C= 0.375kJ/kg.K
Question 10:
A 5.00 g sample of aluminum pellets (specific heat capacity = 0.89 J/°C·g) and a 10.00 g sample of iron pellets (specific heat capacity = 0.45 J/°C·g) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 98.1 g of water at 20.1°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.
Answer :
Heat lost by Al.
= 5g x 0.89J/g/°C x (100°C -T) = 445 - 4.45T.
Heat lost by Fe.
= 10g x 0.45J/g/°C x (100°C - T) = 450 - 4.5T.
Heat gained by water.
= 98.1g x 4.18J/g/°C x (T - 20.1°C) = 410T - 8,242.
(445 - 4.45T) + (450 - 4.5T) = 410T - 8242
8242 + 895 = 8.95T + 410T
9137 = 419T
T = 9137/ 419 = 21.8°C Final temperature.
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