A 200 g copper block slides along a horizontal surface. The block has an initial

Question

A 200 g copper block slides along a horizontal surface. The block has an initial speed of 8.0m/s and initial temperature of 12.0 degree C. The block eventually slides to rest due to friction assuming, all the initial kinetic energy is converted to heat energy and 90% of this energy stays in the block, what is the final temperature of the block? C_copper = 0.093 cal/g.c A hotplate has an area of 0.15 m^2 and temperature of 200 degree C the room temperature is 35 degree C how much net power is transmuted to room in one second e = 0.80.

Explanation / Answer

1.

W=KE = KEf-KEi = ½*m*(vf^2-vi^2)=1/2*0.2*(8.0^2-0.0^2) = 6.4 J

Energy stayed in block in form of heat = Q= 0.90W = 0.90*6.4 = 5.76 J = 1.377cal

Q= mccopper(Tf-Ti)

1.377= 0.2*0.093*(Tf – 12)

Tf = 86.03 C

2.

Tf =(200+273.15) K

Ti =(35+273.15) K

P=eA(T22 – T12) = 0.80*5.67*10^-8*0.15*[(200+273.15)^2 – (35+273.15)^2] = 8.77*10^-4 W

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