Question

10.0g of ice, at a temperature of −10.0 °C, is added to a glass of water to cool it down. When all the ice has melted, and the water and the melted ice are mixed thoroughly, the temperature is 6.0 °C. Assuming that there is no heat exchange with the surroundings, and given the above information:

A. State three changes to the ice from when it was added until being mixed with the drink at 6 °C.

B. Calculate the heat removed from the drink to raise the temperature of the ice to 0 °C.

C. Calculate the total heat removed from the drink by the addition of the ice.

Answer 1

Answer:

a is the correct answer

Answer 2

Given:

10.0 g of ice, at a temperature of −10.0 °C, is added to a glass of water to cool it down. When all the ice has melted, and the water and the melted ice are mixed thoroughly, the temperature is 6.0 °C.

To Find:

A. State three changes to the ice from when it was added until being mixed with the drink at 6 °C.  

B. Calculate the heat removed from the drink to raise the temperature of the ice to 0 °C.  

C. Calculate the total heat removed from the drink by the addition of the ice.

Solution:

We know the value of specific heat of ice and water and latent heat of ice:

$\mathbf{Specific \ heat \ of\ ice (c_{pI})= 2.108 \ \dfrac{J}{g^{\circ}C}}$

$\mathbf{Specific \ heat \ of\ water (c_{pw})= 4.187 \ \dfrac{J}{g^{\circ}C}}$

$\mathbf{Latent \ heat \ of\ ice (L)= 334 \ \dfrac{J}{g}}$

Now come to part A:

There are first sensible heating of ice from −10.0 °C to 0 °C.

After that there are phase change from ice to water at 0 °C. Which is possible by absorbing latent heat from water.

Third change is sensible heating of water from 0 °C to 6 °C.

Now come to part B:

The heat removed from the drink to raise the temperature of the ice to 0 °C is possible by absorbing sensible heat:

We know the formula of sensible heat:

$\mathbf{Specific \ heat =m\times c_{pI}\times (T_{f}-T_{i})}$

Where:

m = 10 g

$\mathbf{T_{f}=0^{\circ}C}$

$\mathbf{T_{i}=-10^{\circ}C}$

On putting respective value in above equation:

$\mathbf{Specific \ heat =10\times 2.108\times (0-(-10))}$

$\mathbf{Specific \ heat =10\times 2.108\times 10}$

On simplify:

Specific heat = 210.8 J

The heat removed from the drink to raise the temperature of the ice to 0 °C is 210.8 J.

Now come to part C:

The total heat removed from the drink to raise the temperature of the ice to 6 °C is possible by absorbing sensible heat of ice and water plus latent heat:

$\mathbf{Total \ heat(Q) =m\times c_{pI}\times (T_{f}-T_{i})+L+m\times c_{pw}\times (t_{f}-t_{i})}$

$\mathbf{Total \ heat(Q) =10\times 2.108\times (0-(-10))+334+10\times 4.187\times (6-0)}$

On simplify:

$\mathbf{Total \ heat(Q) =210.8+334+251.22}$

Total heat (Q) = 796.02 J

The total heat removed from the drink by the addition of the ice is 796.02 J.