Part IV. Specific Heat Capacity of an Unknown Metal 51. Volume of water 100 100
ID: 575674 • Letter: P
Question
Part IV. Specific Heat Capacity of an Unknown Metal 51. Volume of water 100 100 52. Calculate the mass of water 1.38 20-9 53. Mass of unknown metal 54. Initial temperature of water in calorimeter T 55. Initial temperature of metal before placing in calorimeter, T, 56. Final temperature of the water, T 57. Final temperature of the metal, T 58. Calculate the specific heat capacity of the metal TA-C "C 305 59. Calculate the heat capacity of the unknown metal 60. Based on your calculated specific heat capacity, what unknown metal is the washer most likely made of? 61. What is the percent error between your calculated specific heat capacity and the specific heat capacity of the metal you identified above reported in the literature? What are some of the possible sources of error in this experiment. If the metal is not pure, skip the %calculation part % errorExplanation / Answer
Ans. #58. Step 1: Heat absorbed by water during increase in temperature:
q = m s dT - equation 1
Where,
q = heat required
m = mass of solution
s = specific heat of water = 4.184 J g-10C-1
dT = Final temperature – Initial temperature = Tf - Ti = (30.50C -22.9) = 7.60C
# Putting the values in above equation-
q = 100.0 g x (4.184 J g-10C-1) x 7.60C = 3179.84 J
Therefore,
Total amount of heat gained by water = 3179.84 J
# Step 2: The total amount of “heat gained by water during its temperature increase from 22.90C to 30.50C” must be equal to the amount of “heat lost by metal sample while it cools from 99.20C to 30.50C”.
So,
Total heat lost by metal = -3179.84 J
(Note: The -ve sign indicates loss of heat).
# Let the specific heat of metal sample be s.
Putting the values for metal sample in equation 1-
-3179.84 J = 81.88 g x s x (30.5 – 99.2)0C
Or, s = 3179.84 J / (5625.156 g 0C)
Hence, s = 0.565 J g-10C-1
Therefore, specific heat of metal = s = 0.565 J g-10C-1
#59. Heat capacity of metal = Specific heat x mass of sample
= 0.565 J g-10C-1 x 81.88 g
= 46.2622 J 0C-1
#60. The metals with a similar value of specific heat are-
Scandium – 0.57 J g-10C-1
Titanium – 0.54 J g-10C-1
# Because specific heat is not a unique property of metals, it is advised to see the values in your reference book, and match the given values with the options provided. Since, no option list is provided, it would be impossible to predict the correct metal used.
#61. % error not possible unless the metal is determined.
However, for explanation, we proceed assuming the metal sample is titanium.
# % error = [(Experimental value – Theoretical value) / Theoretical vlaue] x 100
= [(0.565 J g-10C-1 - 0.54 J g-10C-1) / 0.54 J g-10C-1] x 100
= 4.63 %
# Possible sources of error:
I. Human errors while taking mass and temperature measurements.
II. Cooling of hot metal while it is being transferred into cold water.
III. Recording temperature before thermal equilibrium is established. Etc.