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Asked in JEE Main 25th June 2nd Shift 2022 · Latent heat and change of state
Given: m_Cu=5.0 kg=5000 g, c=0.39 J g⁻¹ ° C⁻¹, L=335 J g⁻¹.
Idea: the ice block is large, so it stays at 0 °C and the copper gives up all the heat it can, cooling the full 500 °C down to 0 °C.
Heat released: Q=m_CucΔ T=5000×0.39×500=9.75×10⁵ J.
All of it goes into latent heat, since the melt water stays at 0 °C: m_ice=Q/L=(9.75×10⁵)/(335).
m_ice=2910 g≈2.9 kg.
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