Respuesta :
Answer : The final temperature of the mixture is [tex]91.9^oC[/tex]
Explanation :
First we have to calculate the mass of water.
Mass = Density × Volume
Density of water = 1.00 g/mL
Mass = 1.00 g/mL × 180 cm³ = 180 g
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] = specific heat of hot water (liquid) = [tex]4.18J/g^oC[/tex]
[tex]c_2[/tex] = specific heat of ice (solid)= [tex]2.10J/g^oC[/tex]
[tex]m_1[/tex] = mass of hot water = 180 g
[tex]m_2[/tex] = mass of ice = 20 g
[tex]T_f[/tex] = final temperature of mixture = ?
[tex]T_1[/tex] = initial temperature of hot water = [tex]97^oC[/tex]
[tex]T_2[/tex] = initial temperature of ice = [tex]0^oC[/tex]
Now put all the given values in the above formula, we get
[tex](180g)\times (4.18J/g^oC)\times (T_f-97)^oC=-(20g)\times 2.10J/g^oC\times (T_f-0)^oC[/tex]
[tex]T_f=91.9^oC[/tex]
Therefore, the final temperature of the mixture is [tex]91.9^oC[/tex]
