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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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If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain afte
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Answer:

m=0.127g

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g

Best regards.

6 0
3 years ago
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