JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Principle of Calorimetry

Principle of Calorimetry

Category : JEE Main & Advanced

Calorimetry means 'measuring heat'.

When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from body at higher temperature to a body at lower temperature till both acquire same temperature. The body at higher temperature releases heat while body at lower temperature absorbs it, so that

Heat lost = Heat gained

i.e. principle of calorimetry represents the law of conservation of heat energy.

(1) Temperature of mixture \[({{\theta }_{mix}})\] is always \[\ge \] lower temperature \[({{\theta }_{1}})\] and \[\le \] higher temperature \[({{\theta }_{H}})\], i.e., \[{{\theta }_{L}}\le {{\theta }_{mix}}\le {{\theta }_{H}}\].

It means the temperature of mixture can never be lesser than lower temperatures (as a body cannot be cooled below the temperature of cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of heating body). Furthermore usually rise in temperature of one body is not equal to the fall in temperature of the other body though heat gained by one body is equal to the heat lost by the other.

(2) Mixing of two substances when temperature changes only : It means no phase change. Suppose two substances having masses \[{{m}_{1}}\] and \[{{m}_{2}}\], gram specific heat \[{{c}_{1}}\] and \[{{c}_{2}}\], temperatures \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] \[({{\theta }_{1}}>{{\theta }_{2}})\] are mixed together such that temperature of mixture at equilibrium is \[{{\theta }_{mix}}\]

Hence, Heat lost = Heat gained

\[\Rightarrow \] \[{{m}_{1}}{{c}_{1}}({{\theta }_{1}}-{{\theta }_{mix}})={{m}_{2}}{{c}_{2}}({{\theta }_{mix}}-{{\theta }_{2}})\]\[\Rightarrow \]\[{{\theta }_{mix}}=\frac{{{m}_{1}}{{c}_{1}}{{\theta }_{1}}+{{m}_{2}}{{c}_{2}}{{\theta }_{2}}}{{{m}_{1}}{{c}_{1}}+{{m}_{2}}{{c}_{2}}}\]

Temperature of mixture in different cases

Condition Temperature of mixture
If bodies are of same material i.e. \[{{c}_{1}}={{c}_{2}}\] \[{{\theta }_{mix}}=\frac{{{m}_{1}}{{\theta }_{1}}+{{m}_{2}}{{\theta }_{2}}}{{{m}_{1}}+{{m}_{2}}}\]
If bodies are of same mass \[{{m}_{1}}={{m}_{2}}\] \[{{\theta }_{mix}}=\frac{{{\theta }_{1}}{{c}_{1}}+{{\theta }_{2}}{{c}_{2}}}{{{c}_{1}}+{{c}_{2}}}\]
If \[{{m}_{1}}={{m}_{2}}\] and \[{{c}_{1}}={{c}_{2}}\] \[{{\theta }_{mix}}=\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\]

(3) Mixing of two substances when temperature and phase both changes or only phase changes: A very common example for this category is ice-water mixing.

Suppose water at temperature \[{{\theta }_{W}}^{o}C\] is mixed with ice at \[{{0}_{i}}^{o}C,\] first ice will melt and then it's temperature rises to attain thermal equilibrium. Hence;  Heat given = Heat taken

\[\Rightarrow \] \[{{m}_{W}}{{C}_{W}}({{\theta }_{W}}-{{\theta }_{mix}})={{m}_{i}}{{L}_{i}}+{{m}_{i}}{{C}_{W}}({{\theta }_{mix}}-0{}^\circ )\]

\[\Rightarrow \] \[{{\theta }_{mix}}=\frac{{{m}_{W}}{{\theta }_{W}}-\frac{{{m}_{i}}{{L}_{i}}}{{{C}_{W}}}}{{{m}_{W}}+{{m}_{i}}}\]

(i) If \[{{m}_{W}}={{m}_{i}}\] then \[{{\theta }_{mix}}=\frac{{{\theta }_{W}}-\frac{{{L}_{i}}}{{{C}_{W}}}}{2}\]

(ii) By using this formulae if \[{{\theta }_{mix}}<{{\theta }_{i}}\] then take \[{{\theta }_{mix}}=0{}^\circ C\]

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