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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: tricky enough!!!!!!!!!
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karthik_karthik9920 (216)

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guys.. i no the way im supposed 2 solve these bt having trouble getting answer...so plz. solve nd tell mah..these r frm hcv. vol2..caloremetry
 
6] a cube of iron [density 8000kg per m3  specific heat capacity 470j/kg/k ] is heated 2 a high temp  nd is placed on a large ice block of ice at  273k .. the cube melts the ice belo it, displaces the water nd sinks..in the final eq. position, itz upper surface is just inside . calculate initial temp of cube..neglet radiations...density of ice 900kg/m3  latent heat of fusion..336000j/kg
 
ans..353k
 
 
12]a brick of mass 4kg is dropped into a 1 m deep river frm a ht 2m..assume 80%of PE is converted into thermal energy...find thermal energy in calorie...
 
ans...23cal....
 
 
plz..reply...ill certailnly rate u.....u can answer at least 1...
 
 
 
 
 
    
karthik_karthik9920 (216)

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guys plz..reply!!!!!!!!
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shubham_sachdeva (1876)

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whats the difficulty..??
 
q1.
 
first of all find the volume of water..as cube is just suffiecient to melt the whole piece of ice. so
 
 
8000.VCUBE . 470. = MICE.336000
 
NOW AS BLOCK JUST SINKS SO
 
APPLY CONDN. OF SINKING......THEN ITS EASY MAN...

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spideyunlimited (3555)

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Q2) 1 cal = 4.18 Joules

Potential energy = mgh
= 4kg * 10 * 3m
= 120 Joules

in calories --> 120 / 4.18 = 28.708 Cal approx
80% of this is converted to thermal energy
So
80/100 * 28.708
= 22.96 Calories
( Approx 23 Calories)

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karthik_karthik9920 (216)

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sum 1 ans the fist q also..plz...
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