Question

2. DANS A piece of unknown metal weighs 100.0 g. It is heated to 98.0°C before it was dropped into a calorimeter containing 50.0 g of water at 22.0°C. The final temperature was observed to be 26.4'C. Calculate the specific heat capacity of the metal. Type your answer

Answer 1

We know that after dropping the heated metal in the calorimeter containing water, the metal will lose heat and the water will continue to gain heat untill an equilibrium is reached when the temperature of both the metal and the water becomes equal.

Now heat supplied or gained is given by,

q = m * Cs * $\Delta$ T

Where, m= Mass(in grams), Cs = Specific heat capacity( J oC-1g-1), $\Delta$ T = Rise in temperature.

Now, for metal we have,

q1 = 100g * Cs * (Tfinal- TInitial )

= 100g * Cs * ( 26.4 - 98 )oC

= - 7160Cs -------- (1)

For water, we know that Cs = 4.18 J oC-1g-1 , So,

q2 = 50g * 4.18 * (Tfinal - TInitial )

= 50g * 4.18 * ( 26.4 - 22)oC

= 919.6 J ---------- (2)

Now, this amount q2 is the amount of heat gained by water, hence, this same amount of heat is lost by the metal.

Now, according to the law of conservation of energy,

q1 + q2 = 0

therefore, q1 = -q2 ------------ (3)

Now, from equations (1), (2) and (3), we have,

-7160Cs = -919.6 J

Therefore, Cs = ( -919.6/ -7160) J g-1 oC-1

= 0.128  J g-1 oC-1 Answer.

This is the required specific heat capacity of the metal.

Hope it helps.

Thanks and best wishes.