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

trig 15.3 cos

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.