A 1.4-kg piece of unknown metal has an initial temperature of -265 o C. It is pl

Question

A 1.4-kg piece of unknown metal has an initial temperature of -265 oC. It is placed in 300 g of water whose initial temperature is 2 oC. All of the water freezes, and the system comes to equilibrium at a final temperature of -3 oC. The specific heat of water is c = 4186 J/(kg*oC), its latent heat of fusion is Lf = 334 kJ/kg, and the specific heat of ice is c = 2090 J/(kg*oC).

a.) How much heat (in J) is lost by the liquid water as it cools?

b.) How much heat (in J) is lost by the water as it freezes?

c.) How much heat (in J) is lost by the ice as it cools?

d.) What is the specific heat (in J/kg*oC) of the unknown metal?

Explanation / Answer

a) Mw*Cw*dT1

Q1 = 0.3*((4186*2) = 2511.6 J

b) Q2 = Mwf = 0.3*334000 = 100200 J

c) Q3 = M*cice*dT2 = 0.3*2090*3 = 1881 J

d) Qloss = Qgain = Q1 + Q2 + Q3

M*C*dT = 2511.6 + 100200+ 1881

C = 285.14 J /kg oc

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