Data Unknown Identifier(s) (a) Mass of empty styrofoam cups (b) Mass of cups 100
ID: 532611 • Letter: D
Question
Data Unknown Identifier(s) (a) Mass of empty styrofoam cups (b) Mass of cups 100 mL water (c) Mass of cups 100 mL water hot metal (d) Initial temperature of water in the calorimeter (cups) (e) Temperature of the boiling water bath Final temperature of water added hot metal Calculations (g) Mass of water in cups, mcw.[b-a] (h) Mass of added hot metal, mHM Te-b] li) Temperature change of the water in the calorimeter (AT G) Temperature change of the added hot metal (AT, Tf-Ti x AT Calculate the heat gained by the calorimeter B x AT, (m) Calculate the total heat gained AH (n) Since AH calculate the heat lost Earned by the hot metal AHM -AH (o) Calculate the specific heat capacity of the metal from x AT x C (D) Atomic weight nf metal Metal 2 Metal 1. 2.61 g 10 13 g 102.22 150.47 g ZI, Z 2 1,1 13.34 g 21.12 33.09 2. 7 1.2 16 37.29 33, 32 J 91. Z 1661. 3 60 28 72 166 93 J 6023.72Explanation / Answer
Data Unknown Identifiers
Metal 1
Metal 2
a) Mass of empty Styrofoam cups
8.64 g
9.10 g
b)Mass of cups + 100 mL water
107.48 g
108.22 g
c) Mass of cups +100 mL water + hot metal
150.47 g
141.31 g
d) Initial temperature of water in the calorimeter (cups)
21.4C
21.2C
e) Temperature of the boiling water bath
95.4C
95.6C
f) Final temperature of water + added hot metal
24.2C
25.2C
Calculations
g) Mass of water in cups, mCW, (b – a)
(107.48 – 8.64) g = 98.84 g
(108.22 – 9.10) g = 99.12 g
h)Mass of added hot metal. mHM, (c – b)
(150.47 – 107.48) g = 42.99 g
(141.31 – 108.22) g = 33.09 g
i) Temperature change of the water in the calorimeter (TCW = Tf – Ti(CW))
(24.2 – 21.4)C = 2.8C
(25.2 – 21.2)C = 4.0C
j) Temperature change of the added hot metal (THM = Tf – Ti(HM))
(24.2 – 95.4)C = -71.2C
(25.2 – 95.6)C = -70.4C
k) Calculate the heat gained by the cold water
HCW = mCW*CS*TCW (Cs = 4.18 J/g.C)
(98.84 g)*(4.18 J/g.C)*(2.8C) = 1156.823 J
(99.12 g)*(4.18 J/g.C)*(4.0C) = 1657.864 J
l) Calculate the heat gained by the calorimeter
Hcal = B*TCW (take the value of B from below)
(7.333 J/C)*(2.8C) = 20.5324 J
(7.333 J/C)*(4.0C) = 29.332 J
m) Calculate the total heat gained
Hgained = HCW + Hcal
(1156.823 + 20.5324) J = 1177.3554 J
(1657.864 + 29.332) J = 1687.196 J
n) Since Hgained = -Hlost, calculate the heat lost by the metal
HHM = -Hgained
-(1177.3554 J) = -1177.3554 J
-1687.196 J
o) Calculate the specific heat of the metal
HHM = mHM*CS(HM)*THM
Cs(HM) = HHM/(mHM*THM)
(-1177.3554 J)/(42.99 g).(-71.2C) = 0.3846 J/g.C 0.385 J/g.C
(-1687.196 J)/(33.09 g).(-70.4C) = 0.7242 J/g.C 0.724 J/g.C
Part 1: Determining the calorimetric constant
Data Unknown Identifiers
Trial 1
Trail 2
a) Mass of empty Styrofoam cups
9.64 g
10.41 g
b)Mass of cups + 70 mL water
77.16 g
81.40 g
c) Mass of cups +100 mL water + 30 mL hot water
98.88 g
103.03 g
d) Initial temperature of water in the calorimeter (cups)
21.6C
21.5C
e) Temperature of the boiling water bath
96.6C
93.9C
f) Final temperature of water + added hot water
38.5C
39.1C
Calculations
g) Mass of cool water in cups, mCW, (b – a)
(77.16 – 9.64) g = 67.52 g
(81.40 – 10.41) g = 70.99 g
h)Mass of added hot water, mHW, (c – b)
(98.88 – 77.16) g = 21.72 g
(103.03 – 81.40) g = 21.63 g
i) Temperature change of cool water in the calorimeter (TCW = Tf – Ti(CW)) = (f) – (d)
(38.5 – 21.6)C = 16.9C
(39.1 – 21.5)C = 17.6C
j) Temperature change of added hot water (THW = Tf – Ti(HW)) = (f) – (e)
(38.5 – 96.6)C = -58.1C
(39.1 – 93.9)C = -54.8C
k) Calculate the heat lost by the hot water
HHW = mHW*CS*THW (Cs = 4.18 J/g.C)
(21.72 g)*(4.18 J/g.C)*(-58.1C) = -5274.875 J
(21.63 g)*(4.18 J/g.C)*(-54.8C) = -4954.654 J
l) Calculate the heat gained by the cold water
HCW = mCW*Cs*TCW
(67.52 g)*(4.18 J/g.C)*(16.9C) = 4769.748 J
(70.99 g)*(4.18 J/g.C)*(17.6C) = 5222.592J
m) Calculate the heat gained by the calorimeter
Hcal = -HHW - HCW = -k - l
-(-5274.875 J) – (4769.748 J) = 505.127 J
-(-4954.654 J) – (5222.592 J) = -267.938 J
n) Calculate the calorimetric constant
B = Hcal/TCW = m/i
(505.127/16.9) J/C = 29.889 J/C
(-267.938/17.6) J/C = -15.223 J/C
o) Average calorimeter constant
½*[29.889 + (-15.223)] J/C = 7.333 J/C
Data Unknown Identifiers
Metal 1
Metal 2
a) Mass of empty Styrofoam cups
8.64 g
9.10 g
b)Mass of cups + 100 mL water
107.48 g
108.22 g
c) Mass of cups +100 mL water + hot metal
150.47 g
141.31 g
d) Initial temperature of water in the calorimeter (cups)
21.4C
21.2C
e) Temperature of the boiling water bath
95.4C
95.6C
f) Final temperature of water + added hot metal
24.2C
25.2C
Calculations
g) Mass of water in cups, mCW, (b – a)
(107.48 – 8.64) g = 98.84 g
(108.22 – 9.10) g = 99.12 g
h)Mass of added hot metal. mHM, (c – b)
(150.47 – 107.48) g = 42.99 g
(141.31 – 108.22) g = 33.09 g
i) Temperature change of the water in the calorimeter (TCW = Tf – Ti(CW))
(24.2 – 21.4)C = 2.8C
(25.2 – 21.2)C = 4.0C
j) Temperature change of the added hot metal (THM = Tf – Ti(HM))
(24.2 – 95.4)C = -71.2C
(25.2 – 95.6)C = -70.4C
k) Calculate the heat gained by the cold water
HCW = mCW*CS*TCW (Cs = 4.18 J/g.C)
(98.84 g)*(4.18 J/g.C)*(2.8C) = 1156.823 J
(99.12 g)*(4.18 J/g.C)*(4.0C) = 1657.864 J
l) Calculate the heat gained by the calorimeter
Hcal = B*TCW (take the value of B from below)
(7.333 J/C)*(2.8C) = 20.5324 J
(7.333 J/C)*(4.0C) = 29.332 J
m) Calculate the total heat gained
Hgained = HCW + Hcal
(1156.823 + 20.5324) J = 1177.3554 J
(1657.864 + 29.332) J = 1687.196 J
n) Since Hgained = -Hlost, calculate the heat lost by the metal
HHM = -Hgained
-(1177.3554 J) = -1177.3554 J
-1687.196 J
o) Calculate the specific heat of the metal
HHM = mHM*CS(HM)*THM
Cs(HM) = HHM/(mHM*THM)
(-1177.3554 J)/(42.99 g).(-71.2C) = 0.3846 J/g.C 0.385 J/g.C
(-1687.196 J)/(33.09 g).(-70.4C) = 0.7242 J/g.C 0.724 J/g.C
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