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levacccp [35]
3 years ago
9

A sample of iron absorbs 67.5 j of heat upon which the temperature of the sample increases from 21.5 °c to 28.5 °c. if the speci

fic heat of iron is 0.450 j/g-k, what is the mass (in grams) of the sample?
Chemistry
2 answers:
ikadub [295]3 years ago
8 0

Answer : The mass of the sample is, 21.428 grams.

Solution :

Formula used :

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

Q = heat absorbs = 67.5 J

m = mass of sample = ?

c = specific heat of iron = 0.450J/gK      

\Delta T=\text{Change in temperature}  

T_{final} = final temperature = 28.5^oC=273+28.5=301.5K

T_{initial} = initial temperature = 21.5^oC=273+21.5=294.5K

Now put all the given values in the above formula, we get the mass of the sample.

67.5J=m\times 0.450J/gK\times (301.5-294.5)K

m=21.428g

Therefore, the mass of the sample is, 21.428 grams

asambeis [7]3 years ago
3 0
Q = mcΔθ

67.5 = m x 0.45 x (28.5 - 21.5)

M = 67.5 / 3.15
= 21.4 g
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We can use two equations for this problem.<span>

t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is decay constant.

20 days = 0.693 / λ 
λ   = 0.693 / 20 days        (1) 

Nt = Nο eΛ(-λt)                (2)

Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time taken.
t = 40 days</span>

<span>No = 200 g

From (1) and (2),
Nt =  200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>

</span>Hence, 50.01 grams of isotope will remain after 40 days.

<span>
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