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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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