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.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

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

$(180g)\times (4.18J/g^oC)\times (T_f-97)^oC=-(20g)\times 2.10J/g^oC\times (T_f-0)^oC$

$T_f=91.9^oC$

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