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OlgaM077 [116]
3 years ago
9

42.5 grams of an unknown substance is heated to 105.0 degrees Celsius and then placed into a calorimeter containing 110.0 grams

of water at 24.2 degrees Celsius. If the final temperature reached in the calorimeter is 32.4 degrees Celsius, what is the specific heat of the unknown substance?
Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(° C × g).
Chemistry
1 answer:
iVinArrow [24]3 years ago
5 0

Answer:

The specific heat of the unknown substance is 1.22 J/g.°C.

Explanation:

  • Knowing that:

Heat lost by substance (Qc) = Heat gained by the water (Qw) ,

<em>- (Qc) = (Qw).</em>

<em></em>

  • We can calculate the amount of heat (Qw) gained by water using the relation:

Qw = m.c.ΔT,

where, Qw is the amount of heat released to water (Q = ??? J).

m is the mass of water (m = 110.0 g).

c is the specific heat capacity of solution (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = 32.4°C - 24.2°C = 8.2°C).

<em>∴ Q = m.c.ΔT = </em>(110.0 g)(4.18 J/g.°C)(8.2°C) = <em>3770.36 J.</em>

  • Now, the amount of heat lost by the substance <em>(Qc) = - 3770.36 J.</em>

(Qc) = m.c.ΔT,

where, Qc is the amount of heat lost by substance (Qc = - 3770.36 J).

m is the mass of water (m = 42.5 g).

c is the specific heat capacity of solution (c = ??? J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = 32.4°C - 105.0°C = -72.6°C).

∴ (- 3770.36 J) = (42.5 g)(c)(-72.6°C).

∴ c = (- 3770.36 J)/(42.5 g)(-72.6°C) = 1.222 J/g.°C.

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For first-order kinetics, we have:

<em>Rate Law:                                        rate= k[A]</em>

<em>Concentration -Time Equation:      ln[A]=ln[A]o - kt</em>

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To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

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we use:                        [A]=[A]o - Kt

we replace the data:   5 = 100 - K (60)

we clear K:                 K = [100 - 5 ] (mg/L) /60 (s)  = 1, 583 [mg/L.s]

First-order kinetics

we use:                                  ln[A]=ln[A]o - Kt

we replace the data:               ln(5)  = ln(100) - K (60)

we clear K:                                   K = [ln(100) - ln(5)] /60 (s)  = 0,05 [1/s]

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