Question

- Please help me, I don't understand

What is the specific heat of an unknown substance if 100.0 g of it at 200.0 °C reaches an equilibrium temperature of 27.1 °C when it comes in contact with a calorimeter of water. The water weighs 75. g and had an initial temperature of 20.00 °C? (Specific heat of water is 4.18 J/g°C). Show your work

Answer 1

The specific heat of the unknown substance is 0.39 J/g°C.

Explanation:

To solve this problem, we can use the principle of conservation of energy, which states that the heat lost by the unknown substance is equal to the heat gained by the water and the calorimeter. We can express this principle mathematically as:

Q_lost = Q_gained

where Q_lost is the heat lost by the unknown substance, and Q_gained is the heat gained by the water and calorimeter.

We can calculate Q_lost using the formula:

Q_lost = m × c × ΔT

where m is the mass of the unknown substance, c is its specific heat, and ΔT is the change in temperature it undergoes.

We can calculate Q_gained using the formula:

Q_gained = (m_water + m_calorimeter) × c_water × ΔT

where m_water is the mass of the water, m_calorimeter is the mass of the calorimeter, c_water is the specific heat of water, and ΔT is the change in temperature of the water and calorimeter.

Since the system reaches an equilibrium temperature, we can set Q_lost equal to Q_gained and solve for the specific heat of the unknown substance (c).

Here's the calculation:

Q_lost = Q_gained

m × c × ΔT = (m_water + m_calorimeter) × c_water × ΔT

100.0 g × c × (200.0 °C - 27.1 °C) = (75.0 g + 75.0 g) × 4.18 J/g°C × (27.1 °C - 20.00 °C)

Simplifying:

c = [(75.0 g + 75.0 g) × 4.18 J/g°C × (27.1 °C - 20.00 °C)] / (100.0 g × (200.0 °C - 27.1 °C))

c = 0.39 J/g°C

Therefore, the specific heat of the unknown substance is 0.39 J/g°C.

Answer 2

Answer:The specific heat of a substance is defined as the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius (or Kelvin).

To find the specific heat of the unknown substance, we can use the following equation:

Q = m x c x ΔT

where Q is the heat gained or lost, m is the mass of the substance, c is its specific heat, and ΔT is the change in temperature.

In this problem, we know the mass and initial and final temperatures of both the unknown substance and the water, as well as the specific heat of water. We can use this information to calculate the heat gained by the water, which must be equal to the heat lost by the unknown substance:

Heat gained by water = Heat lost by unknown substance

m(water) x c(water) x ΔT(water) = m(substance) x c(substance) x ΔT(substance)

We can plug in the values we know and solve for the specific heat of the unknown substance:

m(water) = 75.0 g

c(water) = 4.18 J/g°C

ΔT(water) = 27.1 °C - 20.00 °C = 7.1 °C

m(substance) = 100.0 g

ΔT(substance) = 200.0 °C - 27.1 °C = 172.9 °C

75.0 g x 4.18 J/g°C x 7.1 °C = 100.0 g x c(substance) x 172.9 °C

Simplifying this equation, we get:

c(substance) = (75.0 g x 4.18 J/g°C x 7.1 °C) / (100.0 g x 172.9 °C)

c(substance) = 0.197 J/g°C

Therefore, the specific heat of the unknown substance is 0.197 J/g°C.

Explanation: