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Question Number 192472 by Mastermind last updated on 19/May/23

$$\mathrm{A}\mathrm{heat}\mathrm{transfer}\mathrm{of}9.0\times {10}^5\mathrm{J}\mathrm{is}\mathrm{required}$$$$\mathrm{to}\mathrm{convert}\mathrm{a}\mathrm{block}\mathrm{of}\mathrm{ice}\mathrm{at}20°\mathrm{C}\mathrm{to}$$$$\mathrm{water}\mathrm{at}15°\mathrm{C},\mathrm{what}\mathrm{was}\mathrm{the}\mathrm{mass}\mathrm{of}$$$$\mathrm{the}\mathrm{block}\mathrm{of}\mathrm{ice}?$$

Answered by MikeH last updated on 19/May/23

$$Q={mc}_i(20-0)+ml+{mc}_w(15-0)$$$$\Rightarrow m=\frac{Q}{20c_i+l+15c_w}$$$$c_i=\mathrm{specific}\mathrm{heat}\mathrm{capacity}\mathrm{of}\mathrm{ice}$$$$c_w=\mathrm{specific}\mathrm{heat}\mathrm{capacity}\mathrm{of}\mathrm{water}$$$$l=\mathrm{laten}\mathrm{heat}\mathrm{of}\mathrm{fusion}\mathrm{of}\mathrm{ice}$$

Commented by Mastermind last updated on 20/May/23

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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