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Question: \( 45g \) of water at \( {{50}^{\circ }}C \) in a beaker is cooled when \( 50g \) of copper at \( {{...

45g45g of water at 50∘C{{50}^{\circ }}C in a beaker is cooled when 50g50g of copper at 18∘C{{18}^{\circ }}C is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is 0.39Jg−1K−10.39J{{g}^{-1}}{{K}^{-1}} and that of water is 4.2Jg−1K−14.2J{{g}^{-1}}{{K}^{-1}} . State the assumption used

Explanation

Solution

Hint : Zeroth law of thermodynamic states that if two objects having different temperatures are kept in content then the transfer of energy between those objects take place until both objects reach the same temperature.

Complete Step By Step Answer:
We are given mass of water mw=45g{{m}_{w}}=45g initial temperature of water Tw=50∘C{{T}_{w}}={{50}^{\circ }}C
Mass of copper mc=50g{{m}_{c}}=50g initial temperature of copper Tc=18∘C{{T}_{c}}={{18}^{\circ }}C
Specific heat capacity of copper sc=0.39J/0/k{{s}_{c}}=0.39J/0/k
Specific heat capacity of water sw=4.2Jg−1k−1{{s}_{w}}=4.2J{{g}^{-1}}{{k}^{-1}}
Let assume final equilibrium temperature = T
Here because water is at high temperature so it will lose heat and copper will gain energy. Heat transferred by water to achieve final temperature Qw=mwsw(50−T){{Q}_{w}}={{m}_{w}}{{s}_{w}}\left( 50-T \right) and heat energy required to copper to attain temperature Qc=mcsc(T−18){{Q}_{c}}={{m}_{c}}{{s}_{c}}\left( T-18 \right)
Now the energy lost by water will be equal to energy gain by copper.
Qw=Qc{{Q}_{w}}={{Q}_{c}}
mwsw(50−T)=mcsc(T−18){{m}_{w}}{{s}_{w}}\left( 50-T \right)={{m}_{c}}{{s}_{c}}\left( T-18 \right)
we can also write it as ⇒50−TT−18=(mcmw)(scsw)\Rightarrow \dfrac{50-T}{T-18}=\left( \dfrac{{{m}_{c}}}{{{m}_{w}}} \right)\left( \dfrac{{{s}_{c}}}{{{s}_{w}}} \right)
by putting the values ⇒50−TT−18=(5045)(0.394.2)\Rightarrow \dfrac{50-T}{T-18}=\left( \dfrac{50}{45} \right)\left( \dfrac{0.39}{4.2} \right)
By solving we get T=47∘CT={{47}^{\circ }}C .

Note :
Unit of specific heat is given in Kelvin terms so we should use the temperature in Kelvin in the formula. But when there is a difference of two temperatures then it is not necessary to convert them into Kelvin first because ΔT\Delta T in degree celsius and ΔT\Delta T in Kelvin will be the same.