Question

A hot solid of mass 60g at ${{100}^{\circ }}C$ is placed in 150g of water at ${{20}^{\circ }}C$. The final steady temperature recorded is ${{25}^{\circ }}C$. Calculate the specific heat capacity of the solid.
[Specific heat capacity of water =4200 $Jk{{g}^{-1}}^{\circ }{{C}^{-1}}$].

Answer 1

Hint: This problem is based on principle of calorimetry by which we can calculate specific heat capacity of substance, to solve this problem we will discuss about specific heat capacity of a substance and how the data gathered from the calorimetric experiment used to calculate specific heat capacity.

Formula used: Principle of calorimetry:
Heat given = heat loss
$\Rightarrow {{m}_{1}}{{c}_{1}}\Delta {{Q}_{1}}={{m}_{2}}{{c}_{2}}\Delta {{Q}_{2}}$

Complete step by step answer:
Calorimetry determines the change in energy of the system by measuring the heat exchanged of substance with the surroundings or we can say by calorimetry we can measure total heat transferred to or from an object.
Given, mass of Solid ${{m}_{1}}=60g$, initial temperature of solid ${{T}_{1}}={{100}^{\circ }}C$, mass of water ${{m}_{2}}=150g$ ,initial temperature of water ${{T}_{2}}={{20}^{\circ }}C$ and final temperature $T={{25}^{\circ }}C$.
Now by applying principle of calorimetry we have,
$\Rightarrow {{m}_{1}}{{c}_{1}}\Delta {{Q}_{1}}={{m}_{2}}{{c}_{2}}\Delta {{Q}_{2}}$
${{c}_{1}}$ specific heat of solid and ${{c}_{2}}$specific heat of water $4.2J{{g}^{-1c}}^{\circ }{{C}^{-1}}$
Now from above equation we have,
$\Rightarrow 60\times {{c}_{1}}\times (100-25)=150\times 4.2\times (25-20)$
$\Rightarrow 60\times {{c}_{1}}\times 75=150\times 4.2\times 5$
$\Rightarrow {{c}_{1}}=\dfrac{42}{60}=0.7J{{g}^{-1}}^{\circ }{{C}^{-1}}$

$\therefore $ The specific heat capacity of solid by the above conclusions will be equal to $0.7J{{g}^{-1}}^{\circ }{{C}^{-1}}$

Additional information: Heat is a form of energy so it can be transferred from one to another so, the higher the temperature higher the energy it possesses, now the specific heat capacity of a substance is defined as the amount of heat required to change the temperature of substance of one gram by one degree Celsius.

Note: In many cases students get confused between heat capacity and specific heat capacity, specific heat capacity is heat needed to increase temperature by one degree Celsius but heat capacity is the ratio of energy provided to a substance and the corresponding increase in temperature.