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A piece of metal is heated to a temperature of 50.0°C and then placed in a calorimeter containing 50.0 g of water at 27.0°C. The water temperature increases to 35.0°C. How many joules of heat were transferred from the metal to the water? (Cwater = 4.18J/g•°C)

Sagot :

Answer: 1672 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed by water = ?  

c = heat capacity of water = [tex]4.18 J/g^0C[/tex]

Initial temperature of the water = [tex]T_i[/tex] = [tex]27^0C[/tex]

Final temperature of the water  = [tex]T_f[/tex]  = [tex]35^0C[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=(35.0-27.0)^0C=8.0^0C[/tex]

Putting in the values, we get:

[tex]Q=50.0g\times 4.18J/g^0C\times 8.0^0C=1672J[/tex]

As heat absorbed by water is equal to the heat released by metal, the Joules of heat transferred are 1672 Joules

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