Thermodynamics, heat in a copper cube transfer to water

AI Thread Summary
A copper cube with a side of 2.0 cm is placed in an insulated container with 1.0 kg of water, causing the water's temperature to rise from 5°C to 7°C. The heat absorbed by the water is calculated to be 8360 J, based on its specific heat capacity. The mass of the copper cube is determined to be 0.0712 kg. Using the heat transfer equation, the change in temperature of the copper is found to be 305°C. The original temperature of the copper cube can be calculated by adding the final temperature of the mixture (7°C) to this change, resulting in an original temperature of 312°C.
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Homework Statement



When a copper cube of side of 2.0 cm is immersed into a perfectly insulated container filled with 1.0 kg of water at 5 degree celcius , the temperature of water rises to 7 degree celcius . Assuming that no heat is lost to the surrounding , calculate the original temperature of the cube .

Given density of copper = 8900 kg / m^3 , specific heat capacity of water = 4180 J/kg/K and specific heat capacity of copper = 385 J/kg/K

Homework Equations





The Attempt at a Solution



from Q=mc theta = 1(4180)(2) = 8360 J and this is the heat supplied to the water

mass of copper = (8900)(8 x 10^(-6)) = 0.0712 kg

8360=0.0712(385) d theta

d theta = 305 degree celcius but this is the change in temperature , how do i find the original temperature from there .
 
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Final temperature of the mixture is 7 degrees
Change in temperature = Original temperature of the copper - final temperature of the mixture.
 
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