JAMB - Physics (2021 - No. 32)
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
स्पष्टीकरण
mc\(\Delta\theta\) = m\(\lambda\)f
2 x 400 (100 - 0) = m(3.3 x 10\(^{5}\))
m = \(\frac{2 \times 400 \times 100}{3.3 \times 10^{5}} = \frac{8}{33}\)kg