Question

In the experiment, it took 300 seconds to increase the temperature of 0.10 kg of the liquid from 25 degrees centigrade to 50 degrees centigrade. During this time, the energy supplied by the heater was 13,600 J and the apparatus was losing it at an average rate of 12 J/s. Assuming that the heat capacity of the container can be ignored, calculate the value of the specific heat capacity of the liquid.

Answer 1

The specific heat capacity of the liquid is calculated to be 4,000 J/kg°C, using the net energy absorbed, the mass of the liquid, and the temperature change during the experiment.

Step-by-step explanation:

To calculate the specific heat capacity of the liquid, we'll use the energy balance concept. The energy supplied by the heater is partly used to increase the temperature of the liquid and partly lost to the surroundings. The formula to use is:

Q = mc∆T

Where:
Q = net energy absorbed by the liquid (Joules)
m = mass of the liquid (kg)
c = specific heat capacity (J/kg°C)
∆T = temperature change (°C)

We know from the experiment that:

• The energy supplied by the heater is 13,600 J.
• Energy lost to the surroundings is 12 J/s * 300 s = 3,600 J.
• Net energy absorbed by the liquid, therefore, is 13,600 J - 3,600 J = 10,000 J.
• The mass of the liquid (m) is 0.10 kg.
• The temperature change (∆T) is 50°C - 25°C = 25°C.

Substituting the values into the formula, we get:

10,000 J = 0.10 kg * c * 25°C

100,000 J/kg = c * 25°C

c = 4,000 J/kg°C

Therefore, the specific heat capacity of the liquid is 4,000 J/kg°C.