
If you wish to warm 100kg of water by 20 degrees Celsius for your bath, how much heat is required?
Answer
542.4k+ views
Hint: You will require an expression for heat required in terms of the given quantities in the question. For that, you could recall the definition of specific heat capacity and hence its expression. Rearranging this you will get the required expression and then you could substitute and get the answer.
Formula used:
Heat required,
$q=Cm\Delta T$
Complete answer:
In the question, we asked that, if we wish to warm 100kg water of 20 degree Celsius in order for us to take bath, then, how much heat is required for the same.
Before answering the question, let me introduce a term that is, specific heat capacity. Heat capacity can be defined as the heat required to change the temperature of a substance by unit degree. Now we have another term that is specific heat capacity where the mass of the substance is specified.
Specific heat capacity is the heat capacity of unit mass of a substance. This is given by the relation,
$C=\dfrac{q}{m\Delta T}$
Where, q is the heat required, m is the mass of the substance and $\Delta T$ is the resultant temperature change.
For water, the value of specific heat capacity is found to be
${{C}_{water}}\approx 4200Jk{{g}^{-1}}{}^\circ {{C}^{-1}}$
Now let us substitute this value along with all the other given values in the question.
$q=Cm\Delta T$
$\Rightarrow q=4200\times 100\times 20$
$\therefore q=8400000J=8.4\times {{10}^{6}}J=8.4MJ$
Therefore, we found that we require 8.4MJ of heat for taking a bath using 100kg of water at 20 degrees Celsius.
Note:
You should understand that the specific heat of a substance varies with temperature. Also, it can have different values for the same substance in its different states. Take the example of ice and water, the specific heat capacity of water (liquid) and ice is very different. When the amount of substance in moles was specified, then, the heat capacity is molar specific heat capacity whose SI unit is $Jmo{{l}^{-1}}{{K}^{-1}}$.
Formula used:
Heat required,
$q=Cm\Delta T$
Complete answer:
In the question, we asked that, if we wish to warm 100kg water of 20 degree Celsius in order for us to take bath, then, how much heat is required for the same.
Before answering the question, let me introduce a term that is, specific heat capacity. Heat capacity can be defined as the heat required to change the temperature of a substance by unit degree. Now we have another term that is specific heat capacity where the mass of the substance is specified.
Specific heat capacity is the heat capacity of unit mass of a substance. This is given by the relation,
$C=\dfrac{q}{m\Delta T}$
Where, q is the heat required, m is the mass of the substance and $\Delta T$ is the resultant temperature change.
For water, the value of specific heat capacity is found to be
${{C}_{water}}\approx 4200Jk{{g}^{-1}}{}^\circ {{C}^{-1}}$
Now let us substitute this value along with all the other given values in the question.
$q=Cm\Delta T$
$\Rightarrow q=4200\times 100\times 20$
$\therefore q=8400000J=8.4\times {{10}^{6}}J=8.4MJ$
Therefore, we found that we require 8.4MJ of heat for taking a bath using 100kg of water at 20 degrees Celsius.
Note:
You should understand that the specific heat of a substance varies with temperature. Also, it can have different values for the same substance in its different states. Take the example of ice and water, the specific heat capacity of water (liquid) and ice is very different. When the amount of substance in moles was specified, then, the heat capacity is molar specific heat capacity whose SI unit is $Jmo{{l}^{-1}}{{K}^{-1}}$.
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