Question

How much heat in calories is needed to raise the temperature of \[125.0\text{ }g\] of Lead from \[17.5{}^\circ C\] to \[41.1{}^\circ C\] ?

Answer 1

Hint :We know that heat is a form of energy that can increase the temperature of liquid, solid or gas. The heat gained at constant volume is when volume doesn’t change and heat gained at constant pressure is when the pressure is constant and doesn’t change.

Complete Step By Step Answer:
The ability to do work is energy. This energy can be measured using multiple different per of units. Calorie and Joule are the units of energy, calorie is in the CGS system and Joule is in the SI system. The relation between calorie and joule is fully proportional. This means that any change in Calorie will change its equivalent value in Joules. That meanswhen the value in terms of calorie increases, its corresponding value in joules also increases at a rate of \[4.2.\] When the value in terms of calorie decreases, its corresponding value in joules also decreases at a rate of \[4.2.\]
Here we know that $q=mc\Delta T$ ………………………………….equation (i)
Where, $q$ is heat energy, $m$ is mass, $c$ is specific heat capacity, and \[\Delta T\] is the change in temperature.
The data here we have is $m=125g,~~~c_{Pb}^{{}}=0.13J/g{}^\circ C,~~~{{T}_{initial}}=17.5{}^\circ C,~~~{{T}_{final}}=42.1{}^\circ C$
Thus, \[\Delta T={{T}_{final}}-{{T}_{initial}}\]
$\therefore \Delta T=42.1{}^\circ C-17.5{}^\circ C=24.6{}^\circ C$
Now we have to substitute the values in equation (i);
$q=\left( 125g \right)\times \left( 0.13J/g{}^\circ C \right)\times \left( 24.6{}^\circ C \right)=400J.$
Now we have to convert joules to calories; since $1J=0.2389cal.$
$400J\times \dfrac{0.2389cal}{1J}=95.6cal$

Note :
Remember that the joule is the unit of energy. Kilojoule is the unit of energy that has \[1000\] joules. The conversion between two units of energy that is calorie to joule and vice versa is crucial in solving problems. The relationship between calories and joule is fully proportional. This means that the calorie is directly proportional to the joule.