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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, whic
Question
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.
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Physics
4 years
2021-07-24T04:21:20+00:00
2021-07-24T04:21:20+00:00 1 Answers
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Answers ( )
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