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# A coffee cup Calorimeter initially contains $125$g of water at a temperature of $24.2^\circ C$. After $10.5$ gm is added to the water at the same temperature, the final temperature becomes $21.1^\circ C$. The heat of solution is-(A) $85{\text{ J/g}}$ (B) $110{\text{ J/g}}$ (C) $270{\text{ J/g}}$ (D) $167{\text{ J/g}}$

Last updated date: 29th Feb 2024
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Hint: First use the formula
$\Rightarrow H = mc\Delta T$ where H is heat energy, c is specific heat and $\Delta T$ is change in temperature. Then use the formula heat of solution=$\dfrac{{{\text{Heat Energy}}}}{{{\text{mass of water added}}}}$ to find the heat of solution. Change the unit from calorie into joule using the value$1{\text{cal = 4}}{\text{.186J}}$to get the answer.

Step-by-Step Solution-
Given, a coffee cup Calorimeter initially contains $125$g of water
Initial Temperature= $24.2^\circ C$
Then $10.5$ gm water is added to the water at the same temperature
Final temperature becomes= $21.1^\circ C$
We have to find the heat of solution.
We know that specific heat is the ratio of the amount of heat required to increase the temperature of an object by one degree to the amount of heat energy required to increase the temperature of water by one degree. So we can write
Heat energy= mass of object× specific heat ×change in temperature
$\Rightarrow H = mc\Delta T$
Now here the total mass of water m= $125 + 10.5 = 135.5$
Specific heat c = $1$
$\Delta T$ = Final temperature- Initial temperature
On putting values we get,
$\Rightarrow \Delta T = \left( {24.2 - 21.1} \right)$
On solving we get,
$\Rightarrow \Delta T = 3.1$
On putting all these values in the formula we get,
$\Rightarrow H = 135.5 \times 1 \times 3.1$
On solving we get,
$\Rightarrow H = 420.05$ cal.
Then heat of solution= $\dfrac{{{\text{Heat Energy}}}}{{{\text{mass of water added}}}}$
On putting the values we get,
Heat of solution= $\dfrac{{420.05}}{{10.5}}{\text{cal}}{{\text{g}}^{{\text{ - 1}}}}$
Heat of solution= $40{\text{cal}}{{\text{g}}^{{\text{ - 1}}}}$
We know that, $1{\text{cal = 4}}{\text{.186J}}$
Then Heat of solution= $40 \times 4.186$ ${\text{J}}{{\text{g}}^{ - 1}}$
On solving we get,
Heat of solution= $167$ ${\text{J}}{{\text{g}}^{ - 1}}$