Respuesta :
Answer : The final temperature of the mixture is, 345.108 K
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] = specific heat of zinc = [tex]0.388J/g.K[/tex]
[tex]C_2[/tex] = specific heat of water = [tex]4.184J/g.K[/tex]
[tex]m_1[/tex] = mass of zinc = 46.1 g
[tex]m_2[/tex] = mass of water = 80.0 g
[tex]T_f[/tex] = final temperature of mixture = ?
[tex]T_1[/tex] = initial temperature of zinc = [tex]18.0^oC=273+18.0=291K[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]75.0^oC=273+75=348K[/tex]
Now put all the given values in the above formula, we get:
[tex]46.1g\times 0.388J/g.K\times (T_f-291)K=-80.0g\times 4.184J/g.K\times (T_f-348)K[/tex]
[tex]T_f=345.108K[/tex]
Therefore, the final temperature of the mixture is, 345.108 K
Answer:
72.11 °C
Explanation:
m₁ = mass of Zn = 46.1 g
m₂ = mass of H₂O = 80 g
c₁ = specific heat of Zn = 0.388 J/(g-K)
c₂ = specific heat of H₂O = 4.184 J/(g-K)
T₁ = initial temperature of Zn = 18 °C
T₂ = initial temperature of H₂O = 75 °C
T = final equilibrium temperature
Using conservation of Heat
Heat gained by Zn = Heat lost by H₂O
m₁ c₁ (T - T₁) = m₂ c₂ (T₂ - T)
(46.1) (0.388) (T - 18) = (80) (4.184) (75 - T)
T = 72.11 °C
