50.0 g of copper was dropped into 250.0mL of H2O at 25.0°C in a coffee cup calorimeter. The copper and H2O reached thermal equilibrium at 27.5°C. Calculate the initial temperature of the copper. Specific heat of water is 4.184 J/g°C and specific heat of copper is 0.385 J/g°C.

Initial temperature of copper = 163.3 °C

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = final temperature - initial temperature

Given data:

Mass of copper = 50 g

Initial temperature of copper = ?

Final temperature = 27.5°C

Initial temperature of water = 25.0°C

Specific heat capacity of copper = 0.385 j/g.°C

Specific heat capacity of water = 4.184 J/g°C

Volume of water = 250.0 mL or 250.0 g

Solution:

Formula:

- Qm  =  +Qw

Now we will put the values in formula.

-50 g × 0.385 j/g.°C × [ 27.5°C -x] = 250 g × 4.184 J/g°C × [ 27.5°C - 25.0°C]

-50 g × 0.385 j/g.°C × [ 27.5°C -x] = 250 g × 4.184 J/g°C × 2.5°C

-19.25 j/°C × [ 27.5°C -x] = 2615 j

27.5°C -x =   2615 j / -19.25 j/°C

27.5°C -x = -135.8°C

x =  27.5°C + 135.8°C

x = 163.3 °C

a

explanation:

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