contestada

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.

Respuesta :

Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celcius. The mass of the remaining ice in the jar will be 630 g.

What are heat gain and heat loss?

Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celcius. While heat loss is inverse to heat gain.

It is given by the formula as ;

Q=mcdt

The given data in the problem is;

Equilibrium temperature = 31.6°C.

mass =173 g,

Mass of ice = 120g

The temperature of the tea =31°C

The specific heat of water =4186 J/kg.

Case 1: Q is the heat gained by ice ;

[tex]\rm Q=mcdT\\\\ \rm Q=0.120 \times 2090({31}^0C-0}^0C)\\\\ \rm Q= 774.8 J[/tex]

Case: 2. Heat Lost by the water

[tex]\rm Q=mCdT\\\\Q= 0.173 \times 4.186\times(31}^0 -31.6}^0)\\\\\rm Q=-434.50J[/tex]

M is the value of the mass of remaining ice in jar =?

h is the enthalpy of fusion of ice per unit mass is 334 J/g.

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

The heat loss by water is equal to the product of enthalpy per unit mass and enthalpy.

H=mh

[tex]\rm (138- M) + 334 =- 434.50J[/tex]

[tex]138+M=-434.50-334\\\\ \rm 138+M=+768.5\\\\\rm Mass=630 g[/tex]

Hence the masses of the remaining ice in the jar will be 630 g.

To learn more about the heat gain refer to the link;

https://brainly.com/question/26268921