Respuesta :
Answer : The final temperature of the mixture will be, [tex]46.7^oC[/tex]
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = [tex]c_2[/tex] = specific heat of liquid water = Same
[tex]m_1[/tex] = mass of liquid water = 100 g
[tex]m_2[/tex] = mass of water = 200 g
[tex]T_f[/tex] = final temperature of mixture = ?
[tex]T_1[/tex] = initial temperature of liquid water = [tex]100^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]20.0^oC[/tex]
Now put all the given values in the above formula, we get
[tex](100g)\times (T_f-100.0)^oC=-[(200g)\times (T_f-20.0)^oC][/tex]
[tex]T_f=46.7^oC[/tex]
Therefore, the final temperature of the mixture will be, [tex]46.7^oC[/tex]
The final temperature will be 46.7 degrees.
Calculation of the final temperature:
Since
specific heat of liquid water = Similar
mass of liquid water = 100 g
mass of water = 200 g
the initial temperature of liquid water = 100
the initial temperature of water = 20
So, the final temperature should be = 46.7
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