Substance -- Specific Heat (J / g-K) N2(g)--1.04 Al(s) --0.90 Fe(s)--0.45 Hg(l)/
ID: 790995 • Letter: S
Question
Substance --Specific Heat (J / g-K)
N2(g)--1.04
Al(s) --0.90
Fe(s)--0.45
Hg(l)/0.14
H2O(l)--4.18
CH4(g)--2.20
CO2(g)--0.84
CaCO3(s)--0.82
1. Which substance on the table requires smallest amount of energy to increase the temperature of 41.5 g of that substance by 11K?
a) Express your answer as a chemical expression. Identify the phase in your answer.
b) Calculate the energy needed for this temperature change.
Express your answer in joules using 2 significant figures
2. The metabolism of glucose yields CO2 (g) and water as products. Energy released in this metabolic process is converted to useful work,"w" with about 68% efficiency.
Calculate the mass of glucose metabolized by a 81.1kg person climbing a mountain with an elevation gain of 1730 m. Assume that the work performed in the climb is four times that required to simply lift 81.1 kg by 1730 m.
Delta H of Glucose -1273.3 kJ/mol
Explanation / Answer
1. (a) The answer is: Hg(l)
Hg(l) requires the least amount of energy since its specific heat is lowest
1. (b) Energy = mass x specific heat x temperature change
= 41.5 x 0.14 x 11
= 63.91 J = 64 J
2. Work = 4 x mass x standard gravity x height
= 4 x 81.1 x 9.807 x 1730
= 5503806 J = 5503.81 kJ
Moles of glucose (100% efficiency) = -work/Delta H
= 5503.81/1273.3 = 4.3225 mol
Moles of glucose (68% efficiency) = 100/68 x 4.3225
= 6.3566 mol
Mass of glucose = moles x molar mass of glucose
= 6.3566 x 180.16
= 1145 g = 1.145 kg (or approximately 1.15 kg)