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im trying to figure out what would happen if 4500 calories of heat are added to a 50 gram ice cube at -10 degrees celsius

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ju

  1. Do you know the specific heat of solid water (ice)? If so, then it will take
    q calories = mass ice x specific heat ice x (0+10) to raise T from -10 C to zero C. Subtract that from 4500 to see how much(if any) heat is left. If some heat is available, then it can go to melting the ice. It takes 80 calories of heat to melt a gram of ice.

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    DrBob222

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