Question

A copper can of mass 120g and containing 150g of water at 27°C is heated for a definite time by means of a small immersion heater. The final temperature being 31°C. The same experiment is repeated with a can containing 13.59g of liquid at the temperature. After heating for the same length of time, it is 46°C. Assuming no heat losses and that the heat capacity of the immersion heater may be neglected, calculate the specific heat capacity of the liquid. (Specific heat of copper = 0.39 J/gK, specific heat of water = 4.18 J/gK)

Answer 1

The specific heat capacity of the liquid cannot be determined with the given data.

Step-by-step explanation:

To calculate the specific heat capacity of the liquid, we can use the equation:

Q = mcΔT

Where Q is the heat energy, m is the mass of the liquid, c is the specific heat capacity of the liquid, and ΔT is the change in temperature. We can write two equations based on the given data:

For the first experiment:

Q = mcΔT

Q = 150g × c × (31°C - 27°C)

Q = 4c

For the second experiment:

Q = mcΔT

Q = 13.59g × c × (46°C - 27°C)

Q = 19.59c

Since the experiments were conducted for the same length of time and neglecting heat losses, the heat energy supplied is the same in both experiments. Therefore, we can set up the following equation:

4c = 19.59c

Solving for c:

19.59c - 4c = 0

15.59c = 0

c = 0

Therefore, the specific heat capacity of the liquid cannot be determined with the given data.