Physics, 05.09.2019 17:10 zahradawkins2007

You have 1.50 kg of water at 28.0∘ in an insulated container of negligible mass. you add 0.700 kg of ice that is initially at -21.0 ∘c. assume that no heat exchanges with the surroundings.
(a) after the thermal equilibrium has been reached, has all the ice melted?
(b) if all of the ice is melted, what is the final temperature of the water in the container? if some ice remains, what is the final temperature of the water in the container, and how much ice remains?

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Answer from: QueenElizabeth353

(a) No

When the water and the ice mix, the amount of heat released by the water is absorbed by the ice. The heat released by the water is "used" by the ice as follows:

1- partially to increase its temperature from -21.0 ∘C to its melting point, 0 ∘C

2- if there is still enough heat, then it is used to melt the ice

3- if there is still enough heat, then it is used to increase the temperature of the "new sample" of water, until the two samples of water are at same temperature (equilibrium)

Let's start with process 1). The heat required for this process is

Q_1=m_i C_i (T_f - T_i )

where

m_i = 0.700 kg is the mass of the ice

$C_i = 2108 J/kg^{\circ}$ is the heat specific capacity of ice

$T_f=0^{\circ}$ is the final temperature of the ice

$T_i = -21.0^{\circ}$ is the initial temperature

Substituting,

$Q_1 = (0.700)(2108)(0-(-21))=3.10\cdot 10^4 J$

When the ice has reached this temperature, then it will continue absorb heat from the water to melt completely. In order to completely melt the ice, the water must have "enough heat" to give off. The maximum heat that the water can give to the ice is when it reaches thermal equilibrium with it, so

Q = m_w C_w (T_f - T_i)

where

m_w = 1.50 kg is the mass of the water

$C_w = 4186 J/kg^{\circ}C$ is the heat specific capacity of water

$T_f=0^{\circ}C$ is the final temperature at equilibrium

$T_i = 28.0^{\circ}C$ is the initial temperature of the water

Substituting,

$Q=(1.5)(4186)(28-0)=1.76\cdot 10^5 J$

Part of this heat is used for process (1), while the rest is used for process (2) (and if there is still heat available, for process 3). However, the heat required to completely melt the ice (process 2) is

$Q_2 = m_{im} \lambda_f$

where

m_i = 0.700 kg is the mass of the ice

$\lambda_f = 3.34\cdot 10^5 J/kg$ is the latent heat of fusion of ice

Substituting,

$Q_2=(0.700)(3.34\cdot 10^5)=2.34\cdot 10^5 J$

We see that Q_2 is larger than Q: this means that not all the ice melts.

(b) 0.266 kg, $0^{\circ} C$

Since we know that not all the ice has melted, we can consider processes 1) and 2) only, and we can write that the heat given off by the water is used partially to heat the ice, and the rest to melt part of the ice:

$Q=Q_1 + \lambda_f m^*$

where m^* is the mass of ice that has melted. Solving for this variable, we find:

$m^* = \frac{Q-Q_1}{\lambda_f}=\frac{1.76\cdot 10^5-3.10\cdot 10^4}{3.34\cdot 10^5}=0.434 kg$

So, the mass of ice remained is

M=0.700-0.434=0.266 kg

and the final temperature at equilbrium is 0 degrees, since not all ice has melted.

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You have 1.50 kg of water at 28.0∘ in an insulated container of negligible mass. you add 0.700 kg of...