Answer to Question #93742 in General Chemistry for Kyla

Question

To raise the temperature of 75.0 g of a particular metal by 1.50 celcius reuires 107 cal of heat. What approximate atomic mass of tge metal? what is the metal?

Answer 1

$q=mCdt$

where $m,c$ and $dt$ represent mass of metal,specific heat capacity and change in temperature respectively.

By substituting the values of the variables above,

$107=75 \times C \times 1.5 \space Cal$

$C=\frac{107 }{75 \times 1.5} \space Cal/gm-\degree C$ $=0.951 \space Cal/gm-\degree C$ $=0.951 \times 4.184 \approx 4 J/gm-\degree C$

The highest specific heat capacity of a metal is 3.56 (in same units) which is for lithium.

So,the metal as illustrated in the question is not practically possible.

This can be water$(C=4.2)$ or some other liquid with specific heat capacity 4.