1. When 400 J of heat is added to 5.6 g of olive oil at 23*C, the temperature increases to 87*C. What is the specific heat of the olive oil?

2. A small rock is heated and placed in a foam cup calorimeter containing 20.0 mL of water at 25.0*C. The water reaches a maximum temperature of 27.2*C. How many joules of heat are released by the rock?

3. How many kilojoules of heat are absorbed when 1.00 L of water is heated from 19*C to 82*C?

1. 1.116 J/g.°C.

2. 184.184 J.

3. 263.718 kJ.

Explanation:

1. When 400 J of heat is added to 5.6 g of olive oil at 23°C, the temperature increases to 87°C. What is the specific heat of the olive oil?

To solve this problem, we can use the relation:

Q = m. c.ΔT,

where, Q is the amount of heat absorbed by olive oil (Q = 400.0 J).

m is the mass of olive oil (m = 5.60 g).

c is the specific heat capacity of lead (c = ? J/g.°C).

ΔT is the temperature difference (final T - initial T) (ΔT = 87 °C - 23 °C = 64.0 °C).

∵ Q = m. c.ΔT

∴ c = Q/m.ΔT = (400.0 J) / (5.6 g) (64.0 °C) = 1.116 J/g.°C.

2. A small rock is heated and placed in a foam cup calorimeter containing 20.0 mL of water at 25.0°C. The water reaches a maximum temperature of 27.2°C. How many joules of heat are released by the rock?

Also, we can use the relation:

Q = m. c.ΔT,

where, Q is the amount of heat absorbed by water (Q = ? J).

m is the mass of water (m = d. V = (1.0 g/mL) (20.0 mL) = 20.0 g).

c is the specific heat capacity of lead (c = 4.186 J/g.°C).

ΔT is the temperature difference (final T - initial T) (ΔT = 27.2 °C - 25.0 °C = 2.2 °C).

∴ Q = m. c.ΔT = (20.0 g) (4.186 J/g.°C) (2.2 °C) = 184.184 J.

3. How many kilojoules of heat are absorbed when 1.00 L of water is heated from 19°C to 82°C?

Also, we can use the relation:

Q = m. c.ΔT,

where, Q is the amount of heat absorbed by water (Q = ? J).

m is the mass of water (m = d. V = (1.0 g/mL) (1000.0 mL) = 1000.0 g = 1.0 kg).

c is the specific heat capacity of lead (c = 4.186 J/g.°C).

ΔT is the temperature difference (final T - initial T) (ΔT = 82.0 °C - 19.0 °C = 63.0 °C).

∴ Q = m. c.ΔT = (1.0 kg) (4.186 J/g.°C) (63.0 °C) = 263.718 kJ.