Question

If 46.1 g Zn at 18.0 C is placed in 80.0 g H2O at 75.0 C, what is the final temperature of the mixture? The specific heat capacities of zinc and water are 0.388 J/g·K and 4.184 J/g·K, respectively

Answer 1

72.11 °C

Step-by-step 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

Answer 2

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.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`C_1` = specific heat of zinc =
`0.388J/g.K`

`C_2` = specific heat of water =
`4.184J/g.K`

`m_1` = mass of zinc = 46.1 g

`m_2` = mass of water = 80.0 g

`T_f` = final temperature of mixture = ?

`T_1` = initial temperature of zinc =
`18.0^oC=273+18.0=291K`

`T_2` = initial temperature of water =
`75.0^oC=273+75=348K`

Now put all the given values in the above formula, we get:

`46.1g* 0.388J/g.K* (T_f-291)K=-80.0g* 4.184J/g.K* (T_f-348)K`

`T_f=345.108K`

Therefore, the final temperature of the mixture is, 345.108 K