Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss (specific heat capacity of copper = 400Jkg\(^{1}\)k\(^{-1}\), latent heat of fusion of ice = 3.3 x 10\(^{5}\)Jkg\(^{-1}\))
\(\frac{8}{33}\)kg
\(\frac{3.3}{80}\)kg
\(\frac{80}{33}\)kg
\(\frac{3.3}{8}\)kg
Explanation
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Heat given off by the copper = MCt
= 2 x 400 x (100 - 0)
Heat absorbed by the ice = ML
= M x 3.3 x 105
Heat lost = Heat gained
∴ M x 3.3 x 105 = 2 x 400 x 100
∴ M = (2 x 400 x 100)/3.3 x 105
= 8/33Kg
mcΔθ = ml
2 x 400 (100 - 0) = m(330000)
m = (2×400×100) / 330000
Jst cancel all zeros out.
=8/33 kg
Option A 
The correct answer is 0.8/3.3 equivalent to 8/33. I guess there is typographic error in the answers
i think the temperature should be converted to kelvin before applying it, because temp used on
deriving specific heat capacity is Kelvin
