Question

How much heat, in joules and in calories, must be removed from 1.75 mol of water to lower its temperature from 25 degree celsius to 15 degree celsius?

Answer 1

Hint:Attempt this question with the help of quantitative relationship between heat transfer and temperature change which tell us about the amount of heat required to lower the temperature from 25$^{\circ }C$ to 15$^{\circ }C$.

Formula used: $Q=mc\Delta T$

Complete step-by-step answer:First we should discuss the role of specific heat:-
“The specific heat is the amount of heat per unit mass required to raise the temperature by 1$^{\circ }C$.”
The specific heat of water in the required units is given below:-
${{c}_{water}}=1.00cal/{{g}^{\circ }}C=4.186J/{{g}^{\circ }}C$
Now, the quantitative relationship between heat added or removed and the change in temperature that consequently takes place is as follows:-
$Q=mc\Delta T$
Where,
 Q = heat transfer
 m = mass of the substance
$~\Delta T$= change in temperature
 c = specific heat (depends on the material and phase)
First let us convert given moles of water to grams so as to use in the above formula:-
=$1.75moles\times \dfrac{18g}{1mole}=31.5g$ of water.
$~\Delta T$= (${{T}_{2}}-{{T}_{1}}$) = (15-25) $^{\circ }C$= -10$^{\circ }C$
Now, let’s substitute all the values in the above formula:-
-Heat in terms of calories:-
\[{{Q}_{calories}}=31.5g\times 1.00\dfrac{cal}{{{g}^{\circ }}C}\times -{{10}^{\circ }}C=-315cal\]
 -Heat in terms of Joules:-
\[{{Q}_{Joules}}=31.5g\times 4.186\dfrac{J}{{{g}^{\circ }}C}\times -{{10}^{\circ }}C=-1318.59J\]
Hence, -315 calories or -1318.59 Joules of heat must be removed from 1.75 mol of water to lower its temperature from 25 degree celsius to 15 degree celsius.

Note:-Always convert the given values into the required units so as to obtain results in desired terms. And also prefer to solve the calculations along with units, so that they can cancel out easily and give you an accurate result.