Question

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 specific heat of iron is 0.450 j/g-k, what is the mass (in grams) of the sample?

Answer 1

Q = mcΔθ

67.5 = m x 0.45 x (28.5 - 21.5)

M = 67.5 / 3.15
= 21.4 g

Answer 2

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