Question

180 cm3 of hot tea at 97 °C are poured into a very thin paper cup with 20 g of crushed ice at 0 °C. Calculate the final temperature of the "ice tea."(Hint: think about two processes: melting the ice into liquid and (maybe) warming the melted ice—now liquid.)

Answer 1

Answer : The final temperature of the mixture is
`91.9^oC`

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.

`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 hot water (liquid) =
`4.18J/g^oC`

`c_2` = specific heat of ice (solid)=
`2.10J/g^oC`

`m_1` = mass of hot water = 180 g

`m_2` = mass of ice = 20 g

`T_f` = final temperature of mixture = ?

`T_1` = initial temperature of hot water =
`97^oC`

`T_2` = initial temperature of ice =
`0^oC`

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

`(180g)* (4.18J/g^oC)* (T_f-97)^oC=-(20g)* 2.10J/g^oC* (T_f-0)^oC`

`T_f=91.9^oC`

Therefore, the final temperature of the mixture is
`91.9^oC`