easy 5 points if you answer this physical science problem?

duppaman u · #1 18-07-2007, 08:03

if you heat a cube of iron in a furnace to 475 k and then drop it into 500l of water in a cylindrical tub, causing the water tempture to rise from 10 c to 100 c, what are the dimensions of the cube? ( the density of iron is 7.87 g/cm3)

Tuncay U · #2 18-07-2007, 08:03

water 500L=500kg=500000grams d=1gr/cm^3 water.
Heat equetion,,,,500000x(100-10)=(iron)x(200-100)
.
90x500000=(iron)x100
45000000=100(iron)
iron=450000 ,,,,,,iron=mc(iron)=mx0.03
450000=mx0.03,,,,,,,m=mass=15000000gr,,Volume=m/d,
Volume=15000000/7.87=1900000cm^3 ::V=a^3
a=1.3 x1000=1300cm.

anilbakshi · #3 18-07-2007, 08:04

heat lost by cube of iron = gained by water
equilibrium temp =100 C
m(iron) * c(iron) [202 - 100] = m (w)*cw [100 - 10]
V(iron) d(iron) c(iron)*102 = V(w) d(w)*cw*90

[V(iron)*7.87cm^3]*0.450*102 =
[500*1000*1 cm^3]*4.186*90

V(iron) = 521463.98 cm^3
volume of cube = x^3 = 521463.98 cm^3
x = side of cube = 80.49 cm

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