Question

Two 20.0-g ice cubes at -11.0 °C are placed into 265 g of water at 25.0 C. Assuming no energy is transferred to or from the surroundings, how do you calculate the final temperature of the water after all the ice melts?

Answer 1

`17.266^\circ C`

Explanation:

Let `T_f` be the final temperature of mixture. The ice absorbs heat during two process

`\text{Ice}\to\text{cooling from}-11^\circ C \ \text{to }0^\circ C \to \text{melting at }0^\circ C\to \text{cooling from }0^\circ C \ \text{to}\ T_f`

As there is no loss of heat to or from the surrounding hence by conservation of energy,

`\text{heat absorbed by ice of 20 gm}=\text{heat rejected by water of 265 gm}`

`M_{\text{ice}}(C_{\text{ice}}\Delta T+LH_{\text{ice}}+C_{\text{water}}\Delta T)=M_{\text{water}}C_{\text{water}}\Delta T`

`20(2.108(0-(-11))+333.55+4.187(T_f-0))=265(4.187)(25-T_f)`

`7134.76+83.74T_f=27738.875-1109.555T_f`

`1193.295T_f=20604.115`

`T_f=\frac{20604.115}{1193.295}`

`=17.266^\circ C`

Hence, the final temperature of mixture (ice +water) will be `T_f=17.266^\circ C`