Question

How much heat does 32.0 g of water absorb when it is heated from 25.0 to 80.0 ${}^\circ C$?

Answer 1

Hint: Before solving this question, we have to first know the formula to find Heat Energy. It is represented by q. The formula is: $q=\,mc\vartriangle T$. Now we can put all the values in the formula and can easily find the required heat energy.

Complete answer:
The temperature of the substance will change by a certain amount when heat energy is added to it. Every material has a different relation between heat energy and temperature and the value that describes the relation of it is the specific heat.
$q=\,mc\vartriangle T$
In this, q is the energy
             m is the mass
              c is the specific heat capacity
             $\vartriangle T$is the change in temperature ($\vartriangle T={{T}_{final}}-{{T}_{initial}}$)
The values that are given in this question are-
Mass = 32.0 g
         = $32\times {{10}^{-3}}$= 0.032 kg
For c, we will use the specific heat concept : Water needs to absorb 4.184 J of heat for the temperature to be increased by 1${}^\circ C$of water. Therefore,
c = 4190 J/kg.K
${{T}_{final}}$= 80.0${}^\circ C$
${{T}_{initial}}$= 25.0 ${}^\circ C$
We have to convert ${}^\circ C$to K
For this conversion We have to add 273.15
K = ${}^\circ C$+ 273.15
${{T}_{final}}$= 80 + 273.15
        =353.15 K
${{T}_{initial}}$= 25 + 273.15
          = 298.15 K
To find $\vartriangle T$, we have to subtract the Initial temperature from the final temperature.
$\vartriangle T={{T}_{final}}-{{T}_{initial}}$
$\vartriangle T$= 353.15 K – 298.15 K = 55 K
The only value that is unknown to us-
q =?
so, $q=\,mc\vartriangle T$
Now, we will put the values of the quantities in the formula:
     $q=\,0.032\times 4.184\times 55$
     $q=\,7374.4\,Joules$
So, 7374.4 Joules of Heat energy is required.

Note:
The specific heat capacity is the amount of heat required to raise the temperature of the substance by 1${}^\circ C$.
The units of all the quantities of the Formula are-
 Heat Energy is (Joules, J)
Mass of a substance is (Kilogram, Kg)
Specific heat is (J/Kg.K)
Temperature is ( Kelvins)