A piece of copper of mass 170g is cooled in a freezer. It is then dropped into water at 0'C causing 4g of water to freeze. Determine the temperature inside the freezer. Please explain the working. The answer to this question is -20. Thankyou
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Hello
The enthalpy of fusion for water is 333,5 J/g . When 4 g water freeze, they give off
4*333,5 J = 1334 J.
This heat was taken up by the copper. To enable the copper to do this, it had to be colder by ∆T°, and then be heated up by ∆T° to 0°C.
The heat to take up was = 1334 J = c(Cu)*m*∆T = 0,381 (J/g*°K)* 170°*∆T
∆T = 1334/(0,381*170) = 20,59°. (c(CU) = 0,381 J/g*°K is the specific heat of copper).
So the copper had to be 20,. °C colder than 0° in order to be able to take up the fusion heat.
Regards
The enthalpy of fusion for water is 333,5 J/g . When 4 g water freeze, they give off
4*333,5 J = 1334 J.
This heat was taken up by the copper. To enable the copper to do this, it had to be colder by ∆T°, and then be heated up by ∆T° to 0°C.
The heat to take up was = 1334 J = c(Cu)*m*∆T = 0,381 (J/g*°K)* 170°*∆T
∆T = 1334/(0,381*170) = 20,59°. (c(CU) = 0,381 J/g*°K is the specific heat of copper).
So the copper had to be 20,. °C colder than 0° in order to be able to take up the fusion heat.
Regards