Calculate the heat absorbed by 200 g ice at $0^{\circ} \mathrm{C}$ to change to water at $60^{\circ} \mathrm{C}$. [Specific heat capacity of ice $=2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$, Specific heat capacity of water $=4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{K}^{-1}$, Specific latent heat of ice $\left.=336000 \mathrm{Jkg}^{-1}\right]$.
Explanation
First, we need to calculate the heat absorbed to change the ice from $0^{\\circ} \\mathrm{C}$ to $0^{\\circ} \\mathrm{C}$, which is Q1 = m * L. Then, we need to calculate the heat absorbed to change the water from $0^{\\circ} \\mathrm{C}$ to $60^{\\circ} \\mathrm{C}$, which is Q2 = m * c * ΔT. Finally, we add Q1 and Q2 to get the total heat absorbed.
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