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:

${17.266}^{\setminus} \circ C$

Explanation:

Let ${T}_{f}$ be the final temperature of mixture. The ice absorbs heat during two process

$\setminus \textrm{I c e} \setminus \to \setminus \textrm{c \infty l \in g \mathfrak{o} m} - {11}^{\setminus} \circ C \setminus \setminus \textrm{\to} {0}^{\setminus} \circ C \setminus \to \setminus \textrm{m e < \in g a t} {0}^{\setminus} \circ C \setminus \to \setminus \textrm{c \infty l \in g \mathfrak{o} m} {0}^{\setminus} \circ C \setminus \setminus \textrm{\to} \setminus {T}_{f}$

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

$\setminus \textrm{h e a t \left\mid \mathmr{and} \right\mid b e d b y i c e o f 20 g m} = \setminus \textrm{h e a t r e j e c t e d b y w a t e r o f 265 g m}$

${M}_{\setminus \textrm{i c e}} \left({C}_{\setminus \textrm{i c e}} \setminus \Delta T + L {H}_{\setminus \textrm{i c e}} + {C}_{\setminus \textrm{w a t e r}} \setminus \Delta T\right) = {M}_{\setminus \textrm{w a t e r}} {C}_{\setminus \textrm{w a t e r}} \setminus \Delta T$

$20 \left(2.108 \left(0 - \left(- 11\right)\right) + 333.55 + 4.187 \left({T}_{f} - 0\right)\right) = 265 \left(4.187\right) \left(25 - {T}_{f}\right)$

$7134.76 + 83.74 {T}_{f} = 27738.875 - 1109.555 {T}_{f}$

$1193.295 {T}_{f} = 20604.115$

${T}_{f} = \setminus \frac{20604.115}{1193.295}$

$= {17.266}^{\setminus} \circ C$

Hence, the final temperature of mixture (ice +water) will be ${T}_{f} = {17.266}^{\setminus} \circ C$