A 51.1g sample of brass is put into a calorimeter (see sketch at right) that contains 250.0g of water. The brass sample starts off at 95.4°C and the temperature of the water starts off at 25.0°C. When the temperature of the water stops changing it's 26.4°C. The pressure remains constant at 1atm. Required:
Calculate the specific heat capacity of brass according to this experiment.

0.415 j/g°C

Explanation:

Given that :

Mass of water, m1= 250.0 g

Temperature of water, T1= 25°C

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

Mass of brass , m2= 51.1g

Temperature of brass = 95.4°C

Specific heat capacity of brass , c2=?

Final temperature = 26.4°C

Heat lost by brass = Heat gained by water

mc(dT) = mc(dT)

51.1 * c * (95.4 - 26.4) = 250 * 4.184 * (26.4 - 25)

51.1 * 69 * c = 250 * 4.184 * 1.4

3525.9c = 1464.4

c = 1464.4 / 3525.9

c = 0.4153265

c = 0.415 j/g°C

Specific heat capacity of Brass for the experiment = 0.415 j/g°C

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol