Question

Two 84.5 g ice cubes are dropped into 30 g of water in a glass. If the water is initially at a temperature of 50 C and if the ice comes directly from the freezer at -300 C, what will be temperature of the drink when the ice and the water reach thermal equilibrium ? How much ice and how much water are present at thermal equilibrium ? Ignore the heat capacity of the glass and heat transferred to and from the environment.

Answer 1

Answer:

final temperature will be 0 degree C

Total amount of ice will be

$m_{ice} = 182 g$

total amount of water

$m_{water} = 17 g$

Explanation:

After thermal equilibrium is achieved we can say that

Heat given by water = heat absorbed by ice cubes

so we will have

Heat given by water to reach 0 degree C

$Q_1 = m_1s_1 \Delta T_1$

$Q_1 = 0.030(4186)(50 - 0)$

$Q_1 = 6279 J$

heat absorbed by ice cubes to reach 0 degree

$Q_2 = m_2 s_2 \Delta T_2$

$Q_2 = (0.169)(2100)(30)$

$Q_2 = 10647 J$

so we will have

$Q_2 > Q_1$

so here we can say that few amount of water will freeze here to balance the heat

$10647 - 6279 = mL$

$m = \frac{10647 - 6279}{335000}$

$m = 13 g$

so final temperature will be 0 degree C

Total amount of ice will be

$m_{ice} = 84.5 + 84.5 + 13$

$m_{ice} = 182 g$

total amount of water

$m_{water} = 30 - 13$

$m_{water} = 17 g$