Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C
If 1.0kg copper block is used, T of the mixture will be = 17.4∘C
If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C
If 25g (0.025Kg) of ice is used, T will be= 147.1∘C
Explanation:
H = mcΘ
heat lost by block = heat gained by water
m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.
0.5*900*(200-20) = m₁*4186*(20-0)
m₁ = 450*180/83270
<em>m₁ = 0.973kg</em>
<em>when 1.0kg of aluminium block is used, the final temperature of the mixture will be </em><em>T</em>
heat lost by block = heat gained by water
1.0*900*(200-T) = 0.973*4186*(T-0)
180000 - 900T = 4073T
4973T = 180000
T = 180000/4973 = 36.2∘C
<em>If 1.0kg copper block is used, T of the mixture will be</em>
heat lost by block = heat gained by water
1.0*387*(200-T) = 0.973*4186*(T-0)
77400 - 387T = 4073T
4460T = 77400
T = 77400/4460 = 17.4∘C
<em>If 100g (0.1kg) of ice at 0∘C is used, T will be</em>
<em>heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C</em>
heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg
0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg
90000 - 450T = 418.6T + 33600
418.6T + 450T = 90000 - 33600
868.6T = 56400
T = 56400/868.6 = 64.9∘C
If 25g (0.025Kg) of ice is used, T will be
0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg
90000 - 450T = 104.65T + 8400
104.65T + 450T = 90000 - 8400
554.65T = 81600
T = 81600/554.65 = 147.1∘C
