Answer :
Let’s break this down step by step:
1. Initial Information:
- Mass of the liquid, (m): 500 g (or 0.5 kg)
- Resistance of the heating coil, (R): 12.5 ohms
- Current passing through the coil, (I): 4 A
- Time taken to reach boiling point, (t_1): 10 minutes (or 600 seconds)
- Additional time after boiling starts, (t_2): 2 minutes (or 120 seconds)
- Boiling point of the liquid: Not specified
- Thermal capacity of the can: 400 J/K
2. Energy Supplied:
- The energy supplied to the liquid is converted into heat.
- Using Ohm’s law, we find the power dissipated in the heating coil: [ P = I^2R ] [ P = (4 , \text{A})^2 \cdot 12.5 , \Omega = 200 , \text{W} ]
3. Energy Used for Heating:
The energy used to raise the temperature of the liquid from its initial
temperature to the boiling point: [ Q_1 = Pt_1 ]
4. Energy Used for Vaporization:
After reaching the boiling point, the remaining energy is used for
vaporization: [ Q_2 = Pt_2 ]
5. Total Energy Used:
The total energy supplied is the sum of (Q_1) and (Q_2): [
Q_{\text{total}} = Q_1 + Q_2 ]
6. Specific Heat Capacity ((c)):
- The specific heat capacity of the liquid can be calculated using: [ Q_1 = mc\Delta T ] where (\Delta T) is the temperature change from initial to boiling point. [ c = \frac{Q_1}{m\Delta T} ]
7. Specific Latent Heat of Vaporization ((L)):
The energy used for vaporization is related to the latent heat: [ Q_2 =
mL ] [ L = \frac{Q_2}{m} ]
8. Calculations:
First, let’s find the total energy:
- [ Q_{\text{total}} = Pt_1 + Pt_2 = 200 , \text{W} \cdot 600 , \text{s} +
200 , \text{W} \cdot 120 , \text{s} = 144000 , \text{J} ]
- Next, we need to find the temperature change:
[ \Delta T = \frac{Q_1}{mc} ] [ \Delta T = \frac{144000 , \text{J}}{(0.5 ,
\text{kg})c} Finally, we can calculate (c) and (L).
9. Results:
- Specific heat capacity, (c): [calculations]
- Specific latent heat of vaporization, (L): [calculations]
Please provide the boiling point of the liquid, and I’ll complete the calculations! :)
