Question

calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 19 °c. assume no heat losses to the surroundings. express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Identify heat - transfer equations

The heat lost by water $Q_{w}=m_{w}c_{w}\Delta T_{w}$, and the heat gained by dry - ice for sublimation $Q_{d}=m_{d}\Delta H_{sub}$. According to the conservation of energy (no heat loss to surroundings), $Q_{w} = Q_{d}$.

Step2: Determine relevant constants

The specific heat capacity of water $c_{w}=4.18\ J/(g\cdot^{\circ}C)$. Let's assume we have $m_{w} = 500\ g$ of water initially at $25^{\circ}C$. The temperature change of water $\Delta T_{w}=25 - 19=6^{\circ}C$. The heat of sublimation of dry - ice $\Delta H_{sub}=571\ J/g$.

Step3: Set up the energy - balance equation

$m_{w}c_{w}\Delta T_{w}=m_{d}\Delta H_{sub}$.

Step4: Solve for the mass of dry - ice $m_{d}$

$m_{d}=\frac{m_{w}c_{w}\Delta T_{w}}{\Delta H_{sub}}$. Substitute $m_{w} = 500\ g$, $c_{w}=4.18\ J/(g\cdot^{\circ}C)$, $\Delta T_{w}=6^{\circ}C$, and $\Delta H_{sub}=571\ J/g$ into the equation:
\[

$$\begin{align*} m_{d}&=\frac{500\ g\times4.18\ J/(g\cdot^{\circ}C)\times6^{\circ}C}{571\ J/g}\\ &=\frac{500\times4.18\times6}{571}\ g\\ &=\frac{12540}{571}\ g\\ &\approx22\ g \end{align*}$$

\]

Answer:

$22\ g$