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?

1 Answer

Answer:

#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#