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Question: A hot solid of mass 60g at \({{100}^{\circ }}C\) is placed in 150g of water at \({{20}^{\circ }}C\)....

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

Explanation

Solution

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
⇒m1c1ΔQ1=m2c2ΔQ2\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 m1=60g{{m}_{1}}=60g, initial temperature of solid T1=100∘C{{T}_{1}}={{100}^{\circ }}C, mass of water m2=150g{{m}_{2}}=150g ,initial temperature of water T2=20∘C{{T}_{2}}={{20}^{\circ }}C and final temperature T=25∘CT={{25}^{\circ }}C.
Now by applying principle of calorimetry we have,
⇒m1c1ΔQ1=m2c2ΔQ2\Rightarrow {{m}_{1}}{{c}_{1}}\Delta {{Q}_{1}}={{m}_{2}}{{c}_{2}}\Delta {{Q}_{2}}
c1{{c}_{1}} specific heat of solid and c2{{c}_{2}}specific heat of water 4.2Jg−1c∘C−14.2J{{g}^{-1c}}^{\circ }{{C}^{-1}}
Now from above equation we have,
⇒60×c1×(100−25)=150×4.2×(25−20)\Rightarrow 60\times {{c}_{1}}\times (100-25)=150\times 4.2\times (25-20)
⇒60×c1×75=150×4.2×5\Rightarrow 60\times {{c}_{1}}\times 75=150\times 4.2\times 5
⇒c1=4260=0.7Jg−1∘C−1\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 0.7Jg−1∘C−10.7J{{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.