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

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

Explanation

Solution

Before solving this question, we have to first know the formula to find Heat Energy. It is represented by q. The formula is: q=mcTq=\,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=mcTq=\,mc\vartriangle T
In this, q is the energy
m is the mass
c is the specific heat capacity
T\vartriangle Tis the change in temperature (T=TfinalTinitial\vartriangle T={{T}_{final}}-{{T}_{initial}})
The values that are given in this question are-
Mass = 32.0 g
= 32×10332\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 1C{}^\circ Cof water. Therefore,
c = 4190 J/kg.K
Tfinal{{T}_{final}}= 80.0C{}^\circ C
Tinitial{{T}_{initial}}= 25.0 C{}^\circ C
We have to convert C{}^\circ Cto K
For this conversion We have to add 273.15
K = C{}^\circ C+ 273.15
Tfinal{{T}_{final}}= 80 + 273.15
=353.15 K
Tinitial{{T}_{initial}}= 25 + 273.15
= 298.15 K
To find T\vartriangle T, we have to subtract the Initial temperature from the final temperature.
T=TfinalTinitial\vartriangle T={{T}_{final}}-{{T}_{initial}}
T\vartriangle T= 353.15 K – 298.15 K = 55 K
The only value that is unknown to us-
q =?
so, q=mcTq=\,mc\vartriangle T
Now, we will put the values of the quantities in the formula:
q=0.032×4.184×55q=\,0.032\times 4.184\times 55
q=7374.4Joulesq=\,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 1C{}^\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)