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