Question

18. A copper calorimeter cup with a mass of 100 g contains 96 g of water at 13°C. If 70 g of a substance at 84°C is dropped into the calorimeter, the temperature increases to 20°C. Find the specific heat capacity of the substance?

Answer 1

Answer: To calculate the specific heat capacity of the substance, we can use the principle of conservation of energy. The heat gained by the substance is equal to the heat lost by the water and the calorimeter cup.

The formula to calculate heat is:

Q = mcΔT

Where:

Q is the heat transferred,

m is the mass of the substance (in this case),

c is the specific heat capacity of the substance,

ΔT is the change in temperature.

Given:

Mass of the calorimeter cup (m_c) = 100 g

Mass of water (m_w) = 96 g

Initial temperature of water (T_iw) = 13°C

Mass of the substance (m_s) = 70 g

Initial temperature of the substance (T_is) = 84°C

Final temperature (T_f) = 20°C

First, we need to calculate the heat lost by the water and the calorimeter cup:

Q_lost = (m_w + m_c) * c_water * ΔT_water

Where c_water is the specific heat capacity of water.

Since the temperature change for the water is from 13°C to 20°C:

ΔT_water = T_f - T_iw = 20°C - 13°C = 7°C

Next, we calculate the heat gained by the substance:

Q_gained = m_s * c_s * ΔT_substance

Where c_s is the specific heat capacity of the substance.

Since the temperature change for the substance is from 84°C to 20°C:

ΔT_substance = T_f - T_is = 20°C - 84°C = -64°C

Now, we equate the heat lost by the water and the calorimeter cup to the heat gained by the substance:

Q_lost = Q_gained

(m_w + m_c) * c_water * ΔT_water = m_s * c_s * ΔT_substance

Substituting the given values, we can solve for c_s, the specific heat capacity of the substance.