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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A 0.040.-kg ice cube floats in 0.200 kg of water in a 0.100-kg copper cup; all are at a temperature of 0°C. A piece of lead at 98°C is dropped into the cup, and the final equilibrium temperature is 12°C. What is the mass of the lead?

To determine
The mass of the lead.

Explanation

Section 1:

To determine: Heat required to melt the ice.

Explanation: Heat required to melt the ice is 13320 J.

Given Info: The mass of the ice cube is 0.040 kg, the mass of water is 0.200 kg, the mass of the copper cup is 0.100 kg, temperature of ice cube, water and the copper cup is 0°C , the temperature of the piece of lead is 98°C , and the final temperature is 12°C .

According to the energy conservation, the conservation equation is,

Qgain=Qlost

  • Qgain is the energy gained by the cold substances.
  • Qhot is the energy lost by the hot substances.

The final temperature of the water is 12°C . So, the ice cube at 0°C melts to form water at 0°C and then its temperature raises to 12°C .

Energy is lost be the lead and the energy is gained by the ice, water and the cup.

Formula the energy required to melt the ice from 0°C to 0°C of water is,

QL=miceLf

  • QL is the energy required to melt the ice.
  • mice is the mass of the ice.
  • Lf is the latent heat of fusion of ice.

Substitute 0.040 kg for mice and 3.33×105J/kg for Lf to find QL .

QL=miceLf=(0.040kg)(3.33×105J/kg)=13320J

Thus, the heat required to melt the ice is 13320 J.

Section 2:

To determine: The amount of heat gained by the ice, water and copper cup.

Answer: Amount of heat gained is 2.584×104J .

Explanation:

Given info: The mass of the ice cube is 0.040 kg, the mass of water is 0.200 kg, the mass of the copper cup is 0.100 kg, temperature of ice cube, water and the copper cup is 0°C , the temperature of the piece of lead is 98°C , and the final temperature is 12°C .

Water from Ice cube at 0°C raises its temperature raises to 12°C .

Formula to calculate the heat gained is,

Qgain=QL+[(mice+mW)cW+mcupcCu](TfTi)

  • mW is the mass of water.
  • CW is the specific heat of water.
  • mcup is the mass of the copper cup.
  • cCu is the specific heat of copper.
  • Ti is the initial temperature.
  • Tf is the final temperature.

Substitute 13320 J for QL , 0.040 kg for mice , 0.200 kg for mW , 4186J/kg°C for cW , 0.100 kg for mcup , 387J/kg°C for cCu , 12°C for Tf , and 0°C for Ti

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