Question

A 100 watt electric heater is inserted into a beaker containing 0.5 kg of a liquid and is switched on for 230 seconds. The temperature of the liquid rises by 10°C. Assume no thermal energy lost to the surroundings, what is the specific heat capacity of the liquid?​

Answer 1

460 J/(kg°C)

Step-by-step explanation:

The specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of one unit of mass of the substance by one degree Celsius.

In this problem, we can use the formula:

Q = mcΔT

where Q is the amount of heat energy transferred, m is the mass of the liquid, c is its specific heat capacity, and ΔT is the change in temperature.

We know that the heater has an output of 100 watts, which means it transfers 100 joules of energy per second. In 230 seconds, it will transfer:

Q = 100 W × 230 s = 23,000 J

The mass of the liquid is given as 0.5 kg, and the temperature change is 10°C. So we can rearrange the formula to solve for c:

c = Q / (mΔT) = 23,000 J / (0.5 kg × 10°C) = 460 J/(kg°C)

Therefore, the specific heat capacity of the liquid is 460 J/(kg°C).

Answer 2

4600J/kgC

Step-by-step explanation:

First use the equation Energy transferred = Power × time to get that the energy transferred to the liquid while heating, ΔQ, is 100 × 230 = 23000J.

Now rearranging the equation ΔQ = mcΔT, we get that the SHC, c, = ΔQ/(mΔT). Subbing in our values we get that c = 23000/(0.5 × 10) = 4600J/kgC.