Question

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?

Answer 1

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)[T_f -(99.9 + 273)K] = (59.8g)(4.184J/gK)[T_f - (24.8+273)K]

(-194.425J/K)T_f + 72501.0825J = (250.2032J/K)T_f - 74510.51J

Putting like terms together;

72501.0825J + 74510.51J = (194.425J/K)T_f + (250.2032J/K)T_f

147011.5925J = (444.6282J/K)T_f

$Tf=\dfrac{147011.5925J}{444.6282J/K}$

T_f = (330.64K - 273)⁰C

= 57.6⁰C