Question #109356

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 in units of °C.

Expert's answer

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"

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