A small coffee cup calorimeter contains 28.0 g of H2O at 19.73 oC. A 2.05 g sample of a metal alloy is heated to 98.88 oC and then placed in the water. The contents of the calorimeter come to a temperature of 21.23 oC. What is the specific heat of lead

1.104 J/g°C

Explanation:

Using Q = m × c × ∆T

Where;

m = mass of substance (g)

c = specific hear capacity (J/g°C)

∆T = change in temperature (°C)

For a colorimeter,

Q(water) = - Q(metal)

m. c. ∆T (water) = - m. c. ∆T (metal)

According to the information provided;

For water:

m = 28.0g

c = 4.184 J/g°C

∆T = (21.23 - 19.73°C)

For the metal:

m = 2.05g

c = ?

∆T = (21.23 - 98.88°C)

m. c. ∆T (water) = - m. c. ∆T (metal)

[28 × 4.184 × (21.23 - 19.73°C)] = -[2.05 × c × (21.23 - 98.88°C)]

[117.152 × 1.5] = -[2.05 × c × (-77.65)]

175.728 = -[-159.1825c]

175.728 = 159.1825c

c = 175.728 ÷ 159.1825

c = 1.104

c = 1.104 J/g°C

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explanation:

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