Q. 3.98 - An Introduction to General, Organic, and Biological Chemistry

Question

A $45 g$ piece of ice at $0^{\circ }C$ is added to a sample of water at $8.0^{\circ }C$. All of the ice melts and the temperature of the water decreases to $0.0^{\circ }C$. How many grams of water were in the sample? 

Step-by-Step Solution

Answer

$450 g$ of water in the sample when ice melts added and the temperature of water decreases to $0.0^{\circ }C$

Step1: Given Information.

The following information's are used to find the gram of water in sample:

Mass of water$=45 g$ 

Initial Temperature$=8^{\circ }C$

freezing temperature$=0^{\circ }C$

Step2: Energy Balance.

By using energy balance between ice and water,

Ice absorbs the heat energy lost by water. Water loses heat energy, which causes it to cool, whereas ice absorbs heat energy, which causes it to melt.

Therefore,

$+m_{ice}\lambda =-m_{\mathrm{water}}{C}_p\Delta T_{\text{water }}$

where,

$m_{ice}$ is mass ice, $\lambda$ is latent heat, $m_{\mathrm{water}}$ is mass of water, $C_p$is specific heat of water and $\Delta T_{\text{water }}$ is change in the temperature of water.

A positive sign indicates that heat energy is being gained, whereas a negative sign indicates that heat energy is being lost.

Step3: Find the gram of water.

Water temperature vary from $8^{\circ }C$ to $0^{\circ }C$

$\Delta T_{\text{water }}=T_{\text{final }}-T_{\text{initial }}$

$$\begin{gathered} =0-8^{\circ }C \\ =-8^{\circ }C \end{gathered}$$

We known,

Latent heat, $\lambda =334 J/g$

Specific heat of water, $C_p=4.18 J/g^{\circ }C$

Substitute the values,$+m_{\text{ice }}\left(\lambda \right)=-m_{\text{water }}\left(C_p\right)\Delta T_{\text{water }}$

$+45( \mathrm{g})\times 334( \mathrm{J}/\mathrm{g})=-m_{\text{water }}\times 4.18\left( \mathrm{J}/\mathrm{g}°\mathrm{C}\right)\times -8\left({}^{\circ }\mathrm{C}\right)$

$$\begin{gathered} 15030 J=m_{\text{water }}\times 33.44 J/g \\ {m}_{\text{water }}=\frac{15030 J}{33.44 J/g} \end{gathered}$$

$$\begin{gathered} =449.462 g \\ =450 g \end{gathered}$$

Therefore, the mass of water is $450 g$