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A copper container (mass: 100 g) holds 200 g of water. Temperature of the container and...

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kibauu | eNoter

Posted May 30, 2013 at 7:18 PM via web

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A copper container (mass: 100 g) holds 200 g of water. Temperature of the container and water is 20 degres celcius

A 150 g piece of copper heated to 80 Celcius is placed in the water. The water stirred thoroughly. After sufficient time, to what temperature does water chage?assume that no heat transferred to or from the environtment. The spesific heat of water is 4,2J/g.K and the spesific heat of copper is 0,4J/g.K

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted May 31, 2013 at 12:59 AM (Answer #1)

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The heat lost by the hot copper will be gained by the cold water and the copper container. It can be expressed as:

`-Q_(copper) = Q_(water) + Q_(contai n er)`

`(-mc Delta T)_(copper) = (mc Delta T)_(water) + (mc Delta T)_(contai n er)`

Given:

Mass of water = 200 g
Mass of container = 100 g
Mass of copper = 150 g

Initial temperature of water = `20 ^o Celsius`
Initial temperature of copper = `80 ^o Celsius`

Specific heat of water = `4.2 (J)/(g^o C)`
Specific heat of copper = `0.4(J)/(g^o C)`

Let x = change in the temperature of water

`(-mc Delta T)_(copper) = (mc Delta T)_(water) + (mc Delta T)_(contai n er)`

`-150*0.4*(x - 80) = 200*4.2*(x-20) + 100*0.4*(x-20)`

`-60*(x-80) = 840*(x-20) + 40*(x-20)`

`-60*(x-80) = (840+40)*(x-20)`

`-60*(x-80) = (880)*(x-20)`

`-60x + 4800 = 880x - 17600`

`-60x - 880x = -17600 - 4800`

`-940x = -22400`

`x = 23.8 ^o Celsius`

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