Question

Suppose you have $600.0g$ of room temperature water $({20.0}^{\circ }C)$ in a thermos. You drop $90.0g$of ice at ${0.00}^{\circ }C$ into the thermos and shut the lid.
(a) What is the equilibrium temperature of the system?
(b) How much ice is left (in grams)?

Answer 1

Hint: As we know that heat is the energy that flows from the high temperature to low temperature, so the heat lost by water will be gained by the ice which will be equal to the latent heat of freezing of ice.

Complete Step by step answer:As we know that heat is the energy that flows from the high temperature to low temperature, so the heat lost by water will be gained by the ice which will be equal to the latent heat of freezing of ice.
We are given with mass of water that is $600.0g$ and the temperature is $({20.0}^{\circ }C)$and we first need to calculate the amount of heat lost and for that we know that heat capacity is given by the formula:
$Q=mc\mathrm{\Delta }T$ where $m$ is the mass, $c$ is calories and $\mathrm{\Delta }T$ is change in temperature.
After we put the values in the formula, we will get:
$Q=600.0\times 1\times ({20}^{\circ }-0^{\circ }C)$
$Q=12000cal$
And again for the ice to melt the heat gained by the ice is equal to $mL(f)$where L is the latent heat so we get,
$$\begin{gathered} Q=90.0\times 1\times ({100}^{\circ }C-{20}^{\circ }C) \\ \Rightarrow Q=7200cal \end{gathered}$$
Therefore, the equilibrium temperature will be $0^{\circ }C$.
(b)Now, since we know that the heat which the water has lost will be equal to the mass of the ice and latent heat of freezing of the ice, so we can calculate the mass as:
$12000cal=m\times 1\times 80ca\lg m^{-1}$
$m=150g$

So the mass of the ice which remained is $150-150=0g$

Note: Whenever the heat energy is supplied to the system usually the temperature increases except during phase transition and rise in temperature is different for different substances and depends on heat capacity. Heat capacity is an extensive property since it depends on the amount of the matter present.