A 505 g piece of copper tubing is heated to 99.9 degrees Celsius & placed in an insulated vessel containing 59.8 g of water at 24.8 degrees Celsius. Assuming that no heat is lost to the vessel, what is the final temperature of the system?
Use the equation;
q =mc"\\Delta"T
According to Law of conservation of energy;
Heat lost by copper tubing = Heat gained by the water
Thus; -(mc"\\Delta"T)Cu = (mc"\\Delta"T)
-(505g)(0.385J/gK)[Tf -(99.9 + 273)K] = (59.8g)(4.184J/gK)[Tf - (24.8+273)K]
(-194.425J/K)Tf + 72501.0825J = (250.2032J/K)Tf - 74510.51J
Putting like terms together;
72501.0825J + 74510.51J = (194.425J/K)Tf + (250.2032J/K)Tf
147011.5925J = (444.6282J/K)Tf
"Tf=\\dfrac{147011.5925J}{444.6282J\/K}"
Tf = (330.64K - 273)0C
= 57.60C
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