Answer to Question #109356 in Mechanics | Relativity for Mark Marksman

Question

A 24 g block of ice is cooled to -88°C. It is added to 538 g of water in an 91 g copper calorimeter at a temperature of 29°C. Find the final temperature. The specific heat of copper is 387 J/kg • °C. The latent heat of fusion of water is 3.33 x 10^5 J/kg and its specific heat is 4186 J/kg • °C.
Answer in units of °C.

Answer 1

Problem:  specific heat of copper is $C_{Cu}=387 J\cdot kg^{-1}\degree C^{-1}$ ;

specific heat of water is $C_{w}=4186 J\cdot kg^{-1}\degree C^{-1}$ ;of ice is $C_{ice}=2100 J\cdot kg^{-1}\degree C^{-1}$ ;  The latent heat of fusion of water is $\lambda=3.33\cdot 10^5 J\cdot kg^{-1}$

Solution: 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 $0\degree C$

(1) $Q_{ice1}=C_{ice}\cdot m_{ice}\cdot \Delta T_{ice}$ where $m_{ice}=0.024 kg$ and $\Delta T_{ice}=0\degree C-(-88)\degree C=88\degree C$

$Q_{ice1}=C_{ice}\cdot m_{ice}\cdot \Delta T_{ice}=2100\cdot 0.024 \cdot 88=4435.2 J$

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

(2) $Q_{ice2}=\lambda\cdot m_{ice}=3.33\cdot 10^5 \cdot 0.024=7992 J$

Third is to warm water of ice block to final temperature

(3) $Q_{ice3}=C_{w}\cdot m_{ice}\cdot T$

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 

(4) $Q_{w}=C_{w}\cdot m_w\cdot(T- T_{init})$ where $m_w=0.538 kg$ , and $T_{init}=29\degree C$. The heat of copper changed by

(5) $Q_{Cu}=C_{Cu}\cdot m_{Cu}\cdot (T-T_{init})$ where $m_w=0.091 kg$

The law of conservation of energy in calorimeter is

(6) $Q_{ice1}+Q_{ice2}+Q_{ice3}+Q_{w}+Q_{Cu}=0$ Substitute (1)-(5) to (6) we find $T$

$T=\frac{(C_{w}\cdot m_w+C_{Cu}\cdot m_{Cu})\cdot T_{init} - Q_{ice1}-Q_{ice2}}{C_{w}\cdot m_w++C_{Cu}\cdot m_{Cu}+C_{w}\cdot m_{ice}}=\frac{(4186\cdot 0.538+387\cdot 0.091)\cdot 29- 4435.2-7992}{2287.3+4186\cdot 0.024}=\\{ \\}=\frac{66331.3 -12427.2 }{2387.8}=22.6 \degree C$

Answer: The final temperature of system is $22.6\degree C$