Answer:
[tex]\text { The temperature of the coffee after the ice melts is } 83.4^{\circ} \mathrm{C}[/tex]
Explanation:
Firstly determine heat absorbed by the ice-cube while it is melting:
[tex]\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 }}[/tex]
[tex]\mathrm{q}_{\mathrm{ice}}=\frac{51}{18.0152}[/tex]
[tex]\mathbf{q}_{\text {ice }}=2.83 \mathrm{kJ}[/tex]
[tex]\text { Now heat emitted by the coffee while ice is melting: } q_{\varepsilon}=-q_{\mathrm{ice}}=-2.83 \mathrm{kJ}[/tex]
Now the temperature of the coffee when the ice has melted:
[tex]\mathrm{q}_{\mathrm{c}}=\mathrm{m}(\Delta \mathrm{T})(\mathrm{C})[/tex]
[tex]\Delta \mathrm{T}=\frac{q_{c}}{m c}[/tex]
[tex]\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}=\frac{q_{c}}{m c}[/tex]
[tex]\mathrm{T}_{\mathrm{f}}=\left(\frac{q_{\mathrm{c}}}{m c}\right)+\mathrm{T}_{\mathrm{i}}[/tex]
[tex]\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}[/tex]
[tex]T_{f}=\frac{-2.83 \times 10^{3}}{501.6}+95[/tex]
[tex]\mathrm{T}_{\mathrm{f}}=-5.64+95[/tex]
[tex]\mathrm{T}_{\mathrm{f}}=89.37^{\circ} \mathrm{C}[/tex]
Since both the substances eventually reach thermal equilibrium then they both must have the same final temperature.
[tex]\text { As } \mathrm{q}_{\mathrm{w}}=-\mathrm{q}_{\mathrm{c}}[/tex]
By solving this, equation achieved is:
[tex]\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}}}[/tex]
[tex]\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)}[/tex]
[tex]\mathrm{T}_{\mathrm{f}}=\frac{0+44827.992}{35.53+501.6}[/tex]
[tex]\mathrm{T}_{\mathrm{f}}=\frac{44827.992}{537.13}[/tex]
[tex]\mathrm{T}_{\mathrm{f}}=83.4^{\circ} \mathrm{C}[/tex]
