Question

When 0.4 kg of brass at ${{100}^{\text{o}}}C$ is dropped into 1 kg of water at ${{20}^{\text{o}}}C$, the final temperature is ${{23}^{\text{o}}}C$. Find the specific heat of brass.
$\left( a \right)307Jk{{g}^{\text{-1}}}^{\,\,\,o}C$
$\left( b \right)407Jk{{g}^{\text{-1}}}^{\,\,\,o}C$
$\left( c \right)507Jk{{g}^{\text{-1}}}^{\,\,\,o}C$
$\left( d \right)607Jk{{g}^{\text{-1}}}^{\,\,\,o}C$

Answer 1

Hint: As there is an action of dropping a brass metal into water so, there will definitely be some changes related to temperature. This is because the brass was at ${{100}^{\text{o}}}C$ means at high temperature which was cooled by dropping it into water with ${{20}^{\text{o}}}C$ to make the metal cool. By the formula of specific heat we will solve the numerical.

Formula used:
$C=cm$ where C is the capacity of heat, s is called a specific heat and m is the mass of the substance and ${{c}_{brass}}{{m}_{brass}}\Delta {{T}_{brass}}={{c}_{water}}{{m}_{water}}\Delta {{T}_{water}}$ where it is clear that left side is for brass and other is water and ${{c}_{water}}=4180Jk{{g}^{-1\,\,\,}}^{\text{o}}C$.

Complete answer:
Specific heat: A specific heat is a specific measurement of heat which is used to increase temperature of any substance like metal (here brass). It does not depend upon mass or volume of any substance but it does depend upon the property of a substance like solid, liquid or gas. We will define its units by $J/k{{g}^{\text{o}}}C$ or $J/kgK$. The specific heat can be measured by the formula $C=cm$. If temperature is also included so the formula becomes $C=cm\Delta T$.
As the heat gained is equal to the heat loss due to the dropping of brass into water, specific heat of water will be the same as specific heat of brass. Thus, we get
${{c}_{brass}}{{m}_{brass}}\Delta {{T}_{brass}}={{c}_{water}}{{m}_{water}}\Delta {{T}_{water}}$
$\Rightarrow {{c}_{brass}}\times 0.4kg\times \left( 100-23 \right)k{{g}^{\text{o}}}C=4180\times 1kg\times \left( 23-20 \right)J$
$\Rightarrow {{c}_{brass}}\times 0.4\times 77k{{g}^{\text{o}}}C=4180\times 3J$
$\Rightarrow {{c}_{brass}}\times \dfrac{4}{10}\times 77k{{g}^{\text{o}}}C=4180\times 3J$
$\Rightarrow {{c}_{brass}}=\dfrac{4180\times 3\times 10J}{77\times 4k{{g}^{\text{o}}}C}$
$\Rightarrow {{c}_{brass}}=\dfrac{125400}{308}Jk{{g}^{\text{-1}}}^{\,\,\,o}C$
$\Rightarrow {{c}_{brass}}=407.14Jk{{g}^{\text{-1}}}^{\,\,\,o}C$
Hence, the correct option is $\left( b \right)407Jk{{g}^{\text{-1}}}^{\,\,\,o}C$.

Note: Here, the degree as Celsius and Kelvin does not matter. We will solve the numerical in Celsius as we will solve it for Kelvin. There are some properties on which specific heat is dependent upon, and they are:
(1) Whenever the temperature, the specific heat also gets changed. Since, the specific heat consists of measurement of heat in it also; this is why it has an impact on specific heat.
(2) If we add or let out a little mass from the actual mass of the substance, it is going to affect the value of specific heat.
(3) What substance we are taking along with what phase like solid, liquid or gas then this will also decide the change in a specific heat as well.