Question

An ice cube of mass 8.5g is added to a cup of coffee, whose temperature is 95 degrees Celcius and which contains 120 g of liquid. Assume the specific heat capacity of coffee is the same as that of water. The heat of fusion of the ice (the heat associated with ice melting) is 6.0 kJ/mol. Find the temperature of the coffee after the ice melts.

Answer 1

Answer:

$\text { The temperature of the coffee after the ice melts is } 83.4^{\circ} \mathrm{C}$

Explanation:

Firstly determine heat absorbed by the ice-cube while it is melting:

$\mathrm{q}_{\mathrm{ice}}=\frac{8.5 \mathrm{gm} \text { ice } \times 1 \mathrm{mol} \text { ice } \times 6 \mathrm{kJ}}{18.0152 \mathrm{gm} \text { ice } \times 1 \mathrm{mol} \text { ice }}$

$\mathrm{q}_{\mathrm{ice}}=\frac{51}{18.0152}$

$\mathbf{q}_{\text {ice }}=2.83 \mathrm{kJ}$

$\text { Now heat emitted by the coffee while ice is melting: } q_{\varepsilon}=-q_{\mathrm{ice}}=-2.83 \mathrm{kJ}$

Now the temperature of the coffee when the ice has melted:

$\mathrm{q}_{\mathrm{c}}=\mathrm{m}(\Delta \mathrm{T})(\mathrm{C})$

$\Delta \mathrm{T}=\frac{q_{c}}{m c}$

$\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}=\frac{q_{c}}{m c}$

$\mathrm{T}_{\mathrm{f}}=\left(\frac{q_{\mathrm{c}}}{m c}\right)+\mathrm{T}_{\mathrm{i}}$

$\mathrm{T}_{\mathrm{f}}=\frac{-2.83 \times 10^{3} \mathrm{J}}{120 \mathrm{gm}\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)}+95^{\circ} \mathrm{C}$

$T_{f}=\frac{-2.83 \times 10^{3}}{501.6}+95$

$\mathrm{T}_{\mathrm{f}}=-5.64+95$

$\mathrm{T}_{\mathrm{f}}=89.37^{\circ} \mathrm{C}$

Since both the substances eventually reach thermal equilibrium then they both must have the same final temperature.

$\text { As } \mathrm{q}_{\mathrm{w}}=-\mathrm{q}_{\mathrm{c}}$

By solving this, equation achieved is:

$\mathrm{T}_{\mathrm{f}}=\frac{\mathrm{m}_{\mathrm{w}} \mathrm{T}_{\mathrm{iv}} \mathrm{C}_{\mathrm{w}}+\mathrm{m}_{\mathrm{c}} \mathrm{T}_{\mathrm{ic}} \mathrm{C}_{\mathrm{c}}}{\mathrm{m}_{\mathrm{w}} \mathrm{C}_{\mathrm{W}}+\mathrm{m}_{\mathrm{c}} \mathrm{C}_{\mathrm{c}}}$

$\mathrm{T}_{\mathrm{f}}=\frac{(8.5 \mathrm{gm})\left(0.0^{\circ} \mathrm{C}\right)\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)+(120 \mathrm{gm})\left(89.37^{\circ} \mathrm{C}\right)\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)}{(8.5 \mathrm{gm})\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)+(120 \mathrm{gm})\left(4.18 \mathrm{J} / \mathrm{gm} .^{\circ} \mathrm{C}\right)}$

$\mathrm{T}_{\mathrm{f}}=\frac{0+44827.992}{35.53+501.6}$

$\mathrm{T}_{\mathrm{f}}=\frac{44827.992}{537.13}$

$\mathrm{T}_{\mathrm{f}}=83.4^{\circ} \mathrm{C}$