Question

In an experiment, 426 g of water is in a copper calorimeter cup of mass 205 g. The cup and the water are at an initial temperature of 10.9 oC. An unknown material with a mass of 361 g at a temperature of 296.1 oC is placed in the water. The system reaches thermal equilibrium at 36.1 oC. What is the specific heat of the unknown material? units = J/kg-C

Answer 1

Ans: Given: Mass of the water (m _water) = 426 g = 0.426 kg.

Initial temperature of water ( T_winitial ) = 10.9 ⁰C.

Unknown material of mass (m _unknown ) = 361 g = 0.361 kg

Temperature of unknown material (T _unknown ) = 296.1 ⁰C.

Equilibrium temperature (T _Equilibrium ) = 36.1 ⁰C.

Specific heat capacity of unknown material (C_U ) = ?

Step 1) Initially water in the calorimeter with initial temperature 10.9 ⁰C. and an unknown material at a 296.1 ⁰C.

Step 2) This unknown material is kept inside the water, then system returns to the equilibrium situation at 36.1 ⁰C.

Therefore heat gained by the water is equal to the heat lost by the unknown material

Q _{W gained} = Q _{U Lost}

( m _water ) x c_w x T = ( m _Unknown ) x c_u x T -------------------- (1)

Where,

m _water - is the mass of water in kg.

m _Unknown - is the mass of the unknown material in kg.

c_w - specific heat capacity of water = 4186 J/kg ⁰C.

cu - specific heat capacity of unknown material.

​​​​​​​T - temperature difference

Therefore equation (1) becomes

( m _water ) x c_w x T = ( m _Unknown ) x c_u x ​​​​​​​T

  0.426 x 4186 x ( T _Equilibrium - T_winitial) = 0.361 x c_u x ( T _unknown - T _Equilibrium)

0.426 x 4186 x (36.1 - 10.9) = 0.361 x c_u x (296.1 - 36.1)

1783.236 x 25.2 = 0.361 x c_u x 260

44937.5472 = 93.86 x c_u

  44937.5472 / 93.86 = c_u

c_u = 478.77 J / kg ⁰C

Therefore specific heat capacity of unknown material is 478.77 J / kg ⁰C