Multiple choice

Ice in a freezer is at $-7^{\circ}C. 100\ g$ of this ice is mixed with $200\ g$ of water at $15^{\circ}C$. Take the freezing temperature of water to be $0^{\circ}C$, the specific heat of ice equal to $2.2\ J/g^{\circ}C$, specific heat of water equal to $4.2\ J/g^{\circ}C$, and the latent heat of ice equal to $335\ J/g$. Assuming no loss of heat to the environment, the mass of ice in the final mixture is closest to

  1. $88\ g$
  2. $67\ g$
  3. $54\ g$
  4. $45\ g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Heat lost by water cooling to 0C: 200 * 4.2 * 15 = 12600 J. Heat gained by ice warming to 0C: 100 * 2.2 * 7 = 1540 J. Remaining heat available to melt ice: 12600 - 1540 = 11060 J. Mass of ice melted: 11060 / 335 = 33.01 g. Remaining ice: 100 - 33.01 = 66.99 g, which is approximately 67 g.

AI explanation

The heat lost by the 200 g of water cooling to 0 degrees Celsius is 200 multiplied by 4.2 multiplied by 15, which equals 12600 Joules. The heat required to bring the 100 g of ice to 0 degrees Celsius is 100 multiplied by 2.2 multiplied by 7, equaling 1540 Joules, leaving 11060 Joules to melt the ice. Dividing 11060 Joules by the latent heat of 335 Joules per gram shows that 33 g of ice melts, leaving 67 g of ice in the final mixture.

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