Answer to Question #199064 in Mechanics | Relativity for rackell

Question

A 20 g block of ice is cooled to −65 ◦C. It is added to 577 g of water in an 80 g copper calorimeter at a temperature of 29◦C. Find the final temperature. The specific heat of copper is 387 J/kg · ◦C and of ice is 2090 J/kg · ◦C . The latent heat of fusion of water is 3.33 × 10^5 J/kg and its specific heat is 4186 J/kg · ◦C . Answer in units of ◦C.

Answer 1

Specific heat of copper is $C_{Cu}=387\ J/(kg\cdot\degree C)$

specific heat of water is $C_{w}=4186\ J/(kg\cdot\degree C)$

specific heat of ice is $C_{ice}=2100\ J/(kg\cdot\degree C)$

he latent heat of fusion of water is $\lambda=3.33⋅10 ^5\ J/kg$

The heat capacity of water seams to be much greater then of the ice, thus we will decide the ice will melt and final temperature of system $T>0\degree C$. Determine the heat capacity wich must be added to block of ice. This consist of three terms. First is the heat capacity to warm ice block to $0\degree C$.

$Q_{ice1}=C_{ice}m_{ice}\Delta T$

$m_{ice}=0.020\ kg$

$\Delta T=0-(-65)=65\ \degree C$

$Q_{ice1}=2100\cdot0.02\cdot65=2730\ J$

Second is the heat to melt ice block at $0\ \degree C$ :

$Q_{ice2}=\lambda m_{ice}=3.33\cdot10^5\cdot0.02=6660\ J$

Third is to warm water of ice block to final temperature:

$Q_{ice3}=C_wm_{ice}T$

The law of conservation of thermal energy in this case says that all this energy (1)-(3) ice gets from cooled water and copper calorimeter. The water lost the amount of heat 

$Q_{w}=C_wm_{w}(T-T_{init})$

$m_{w}=0.577\ kg$ , $T_{init}=29\ \degree C$

The heat of copper changed by

$Q_{w}=C_{Cu}m_{Cu}(T-T_{init})$

$m_{Cu}=0.080\ kg$

The law of conservation of energy in calorimeter is

$Q_{ice1}​ +Q_{ice2}​ +Q_{ ice3}​ +Q_{ w}​ +Q_{ Cu}​ =0$

Then the final temperature of system is

$T=\frac{(C_wm_w+C_{Cu}m_{Cu})T_{init}-Q_{ice1}-Q_{ice2}}{C_wm_w+C_{Cu}m_{Cu}+C_{ice}m_{ice}}=\frac{29(4186\cdot0.577+387\cdot0.08)-2730-6660}{4186\cdot0.577+387\cdot0.08+2100\cdot0.02}=24.7\ \degree C$