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

Question

400 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 50.0°C at the same time.

What is 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

First look at what you need to solve:

Here, they reach the same temperature after 'absorbing' the same amount of energy (heat). So strategy is to use the known mass and specific heat to find the energy, and then use the energy with specific heat to find mass.

4148 J/(kg*C) * 30C (chang in temp of liquid) * 0.4kg = 49776 J

For the second part, ice has to warm to 0C, then melt (go to water), then water warms up

{2060 J/(kg*C) *mice * 10C } +{335 KJ/kg * mice} + {4148J(kg*c) *mice *50C} = 49776J

mice ( 20600 J/kg + 207400 J/kg + 335000 J/kg) = 49776 J

mice 88.4g

Always ask yourself if the answer makes sense. In this case the ice sample has to warm by 10C, absorb energy to undergo phase transition, and then warm 50C. The water only has to warm 30C, so naturally if they arrive at the same temperature, after absorbing the same total energy, the mass of the ice must be much less than the mass of the water.

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