Question

A 30.5 g sample of pure metal at 156.7˚C is added to a 0.700 kg of water (specific heat capacity = 4.180 J/˚C g) at 23.0 ˚C. What is the final temperature of the water if the metal is (a) antimony, specific heat capacity of 0.210 J/˚C g (b) iridium, specific heat capacity of 1.30 J/˚C g? Assume all heat is conserved.

Answer 1

A. Let the final temperature of water be x °C.

Now amount of heat released by antimony,

Q_M = M_metal x Specific heat capacity of metal x change in temperature

= 30.5 x 0.210 x (156.7 - x)

= 1003.66 - 6.405x

And amount of heat absorbed by water,

Q_H2O = Mass of water x specific heat capacity of water x Change in temperature

= 700 x 4.180 x (x - 23)

= 2926 x (x - 23)

= 2926x - 67298

As per principle of Calorimetry,

Q_M = Q_H2O

or, 1003.66 - 6.405x = 2926x - 67298

or, 2926x + 6.405 x = 1003.66 + 67298

or, 2932.405 x = 68301.66

or, x = 23.29 °C

B. Let the final temperature of water be x °C.

Now amount of heat released by iridium,

Q_M = M_metal x Specific heat capacity of metal x change in temperature

= 30.5 x 1.3 x (156.7 - x)

= 6213.16 - 39.65x

And amount of heat absorbed by water,

Q_H2O = Mass of water x specific heat capacity of water x Change in temperature

= 700 x 4.180 x (x - 23)

= 2926 x (x - 23)

= 2926x - 67298

As per principle of Calorimetry,

Q_M = Q_H2O

or, 6213.16 - 39.65x = 2926x - 67298

or, 2926x + 39.65 x = 6213.16 + 67298

or, 2965.65 x = 73511.16

or, x = 24.79 °C