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Mechanical Properties Of Fluids

Question

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A copper ball weighing 400gm is takem from a furnace to 1000gm of water at 20°C. The temperature of water increases to 80°C at equilibrium. Find the temperature of furnace. The specific heat of copper is 0.1 cal/ gm/K and that of water is 1.0 cal/gm/K.

Solution

Weight of the copper ball = 400 g
Weight of water = 1000 gm
Temperature = 20^o C
Increased temperature = 80^o C
Let T be the temperature of furnace.
According to law of calorimetry,

Heat lost by copper = Heat gained by water.
That is,
400 x 0.1 x ( T– 80 ) = 1000 x 1 x 60
$rightwards double arrow$ T - 80 = 1580
$rightwards double arrow$ T = 1580⁰C

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