A jar of tea is placed in sunlight until it

reaches an equilibrium temperature of 31.6°C.

In an attempt to cool the liquid, which has a

mass of 173 g, 120 g of ice at 0.0°C is added.

At the time at which the temperature of the

tea is 31°C, find the mass of the remaining

ice in the jar. The specific heat of water

is 4186 J/kg. ° C. Assume the specific heat

capacity of the tea to be that of pure liquid

water.

Answer in units of g.

The mass left by the jar of tea is 630 g

Explanation:

Solving the problem:

1. Heat gained by the Ice

To find the value of the Q

Q=Heat gained by the Ice

Q=m c delta T

Q=0.120kg x 2090 J/kg. ° C. (31°C-0°C)

Q = 774.8 J

2. Heat Lost by the water

Q=Heat Lost by the water

Q=m c deltaT

Q = 0.173 kg x 4186 J/kg. ° C. (31°C-31.6°C)

Q=-434.50J

The results obtained for the heat is

Ice from 0 degC - > 31degC = Q = 774.8 J

water from 31.6 degC - > 31 degC = Q = 434.50J

3. To find the mass

We need to use enthalpy of fusion of ice which is 334 J/g.

The mass of ice remaining at the end has to be less than 138 g.

Mass = (138 - M) x 334 (enthalpy) = 434.50J (heat loss by water)

Mass = 138-M=-434.50-334

Mass=138-M=768.5

Mass=630 g

The mass left by the jar of tea is 630 g