A 33.62 g sample of metal is heated to 98.6°C in a hot water bath until thermal

Question

A 33.62 g sample of metal is heated to 98.6°C in a hot water bath until thermal equilibrium is reached. The metal is quickly transferred to 50.0 mL (50.0 g) of water at 22.2°C contained in a calorimeter. The thermal equilibrium temperature of the metal sample plus water mixture is 28.3°C. What is the specific heat (smetal) of the metal in J/g°C ?

Answer to 3 significant figures. Do not include unit. HINT: Heat given off by metal (-) = Heat absorbed by water (+) -qmetal = +qwater -(smetal)(mmetal)(incrementTmetal) = (swater)(mwater)(incrementTwater)

Explanation / Answer

Answer :

For water,

Mass m = 50.0 g, sp. heat capacity Cpw= 4.184 J/g oC, Initial Temp Ti = 22.2 oC, Final Tf = 28.3 oC

For Metal,

Mass M = 33.62 g, sp. heat capacity Cpm = ? Initial temp Ti = 98.6 oC, Final temp Tf = 28.3 oC

Now by Conservation of energy law,

Heat given off by metal = Heat absorbed by water

- qmetal = + qwater.

- M x Cpm x (Tf-Ti) = + m x Cpw x (Tf-Ti)

- 33.62 x Cpm x (28.3 - 98.6) = + 50.0 x 4.184 x (28.3 - 22.2)

33.62 x Cpm x 70.3 = 50.0 x 4.184 x 6.1

Cpm = 50.0 x 4.184 x 6.1 / (33.62 x 70.3)

Cpm = 0.54 J/ g.oC

Specific heat capacity of metal is 0.54  J/ g.oC

=================XXXXXXXXXXXXXXXXXXX======================

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.