Answer:

m = 450 g

C = 4.18 J/g/°C

Tinitial = 15°C

Tfinal = 85°Cm = 50.0 g

C = 4.18 J/g/°C

Tinitial = 88.6°C

Tfinal = 87.1°C

ΔT = -1.5°C (Tfinal - Tinitial)

Solve for Qwater:

Qwater = m•C•ΔT = (50.0 g)•(4.18 J/g/°C)•(-1.5°C)

Qwater = -313.5 J (unrounded)

(The - sign indicates that heat is lost by the water)

12.9 gram sample of an unknown metal at 26.5°C is placed in a Styrofoam cup containing 50.0 grams of water at 88.6°C. The water cools down and the metal warms up until thermal equilibrium is achieved at 87.1°C. Assuming all the heat lost by the water is gained by the metal and that the cup is perfectly insulated, determine the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.18 J/g/°C.Qmetal = 313.5 J (use a + sign since the metal is gaining heat)

m = 12.9 g

Tinitial = 26.5°C

Tfinal = 87.1°C

ΔT = (Tfinal - Tinitial )

Solve for Cmetal:

Rearrange Qmetal = mmetal•Cmetal•ΔTmetal to obtain Cmetal = Qmetal / (mmetal•ΔTmetal)

Cmetal = Qmetal / (mmetal•ΔTmetal) = (313.5 J)/[(12.9 g)•(60.6°C)]

Cmetal = 0.40103 J/g/°C

Cmetal = 0.40 J/g/°C (rounded to two significant digits)