500 g of water at 20°C is in a pot on the stove. An unknown mass of ice that is

Question

500 g of water at 20°C is in a pot on the stove. An unknown mass of ice that is originally at -10°C is placed in an identical pot on the stove. Heat is then added to the two samples of water at precisely the same constant rate. Assume that this heat is transferred immediately to the ice or water (in other words, neglect the increase in temperature for the pot). We will also neglect evaporation. The ice melts and becomes water, and you observe that both samples of water reach 65.0°C at the same time.

Solve for the mass of the ice that was originally in the second pot. The specific heat of liquid water is 4186 J/(kg °C), and of solid water is 2060 J/(kg °C). The latent heat of fusion of water is 3.35 x 105 J/kg.

Explanation / Answer

q=mcdeltaT so 0.5*4186*45 = 94185 m = 150.05 g

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