Answer to Question #234438 in General Chemistry for Earl

Question

A 100-g metal at 110°C is placed in a constant-pressure

calorimeter containing 120-g of water at 18°C. If the

specific heat of the metal is 0.57 J/gC°, and the calorimeter

is 300 grams at 18°C initially with an average specific heat

of 0.30 J/gC°. Calculate the final temperature of the water.

Answer 1

Given: Mass of metal = 100 g.

Initial temperature of metal = 110 ^oC

Specific heat capacity of metal = 0.57 J/g.^oC

Mass of water = 120 g

Initial temperature of water = 18 ^oC

Mass of calorimeter = 300 g

Initial temperature of calorimeter = 18 ^oC

And specific heat capacity of calorimeter = 0.30 J/g.^oC.

 The total amount of heat lost by the metal is given by,

=> Q = mCΔT

Where m = Mass of metal = 100 g.

C = Specific heat capacity of metal = 0.57 J/g.^oC

Where m = Mass of metal = 100 g.

C = Specific heat capacity of metal = 0.57 J/g.^oC

And ΔT = change in temperature of metal = final temperature - initial temperature = T - 110 

Hence substituting the values we get,

=> Q = 100 X 0.57 X (T - 110) = 57 T - 6270 

Since all the heat lost by metal is gained by the calorimeter and water.

=> total heat gained by calorimeter and water = - Q of metal = 6270 - 57 T (-ve of heat lost as heat lost will be a negative quantity).

The total heat gained by calorimeter and water is also given by,

=> Q = m_wC_wΔT_w + m_cC_cΔT_c 

Where m_w = Mass of water = 120 g

C_w = specific heat capacity of water = 4.184 J/g.^oC

ΔT_w = change in temperature of water = T - 18 

m_c = Mass of calorimeter = 300 g

C_c = specific heat capacity of water = 0.30 J/g.^oC

ΔT_c = change in temperature of water = T - 18

Hence substituting the values we get,

=> 6270 - 57 T = 120 X 4.184 X (T - 18) + 300 X 0.30 X (T - 18)

=> T = final equilibrium temperature of water = 26.1 ^oC approx.