16. A kettle of heat capacity 400Jkg 'C' contains 2kg of water at a temperature of 30°C. The great heat capacity of water is 4000Jkg °C'. A 2kW heating coil was used to raise the water temperature to 80 °C. If there is no thermal damage to the environment, find the time taken by the coil for this process.

To solve this, we will use the formula for heat transfer and power, as well as the specific heat capacity equation.

### Step 1: Calculating the heat required to raise the temperature of water

The total heat required, \( Q \), to raise the temperature of the water can be calculated using the formula:

\[
Q = mc\Delta T
\]

Where:
- \( m \) is the mass of the water = 2 kg
- \( c \) is the specific heat capacity of water = 4000 J/kg°C
- \( \Delta T \) is the change in temperature = \( 80°C - 30°C = 50°C \)

Substituting the values:

\[
Q = 2 \, \text{kg} \times 4000 \, \text{J/kg°C} \times 50 \, \text{°C}
\]

\[
Q = 400,000 \, \text{J}
\]

### Step 2: Calculating the heat absorbed by the kettle

The heat absorbed by the kettle itself is given by:

\[
Q_{\text{kettle}} = mc\Delta T
\]

Where:
- \( m_{\text{kettle}} \) is the mass of the kettle, which we assume is included in the heat capacity (i.e., 400 J/°C)
- \( c_{\text{kettle}} \) is the kettle’s heat capacity = 400 J/°C
- \( \Delta T \) is the same temperature change = 50°C

Thus,

\[
Q_{\text{kettle}} = 400 \, \text{J/°C} \times 50 \, \text{°C} = 20,000 \, \text{J}
\]

### Step 3: Total heat required

The total heat required is the sum of the heat for the water and the heat for the kettle:

\[
Q_{\text{total}} = Q_{\text{water}} + Q_{\text{kettle}} = 400,000 \, \text{J} + 20,000 \, \text{J} = 420,000 \, \text{J}
\]

### Step 4: Calculating the time required

Power is defined as the rate at which energy is transferred. The formula for power is:

\[
P = \frac{Q}{t}
\]

Where:
- \( P \) is the power of the heating coil = 2 kW = 2000 W
- \( Q \) is the total heat required = 420,000 J
- \( t \) is the time required in seconds

Rearranging the formula to solve for \( t \):

\[
t = \frac{Q}{P} = \frac{420,000 \, \text{J}}{2000 \, \text{W}} = 210 \, \text{seconds}
\]

### Conclusion:

The time taken by the coil to raise the water temperature to 80°C is **210 seconds** (or **3 minutes and 30 seconds**).

---

Would you like further details or have any questions?

Here are some related questions you might find interesting:
1. How would the time change if the kettle's heat capacity increased?
2. How would you calculate the energy loss if thermal damage to the environment occurred?
3. What if the heating coil had a power of 1.5 kW instead of 2 kW?
4. How does the specific heat capacity of water affect the overall heat requirement?
5. What happens to the time if the initial temperature of the water is lower?

**Tip:** Always convert units to the SI system (e.g., kW to W) to avoid calculation errors.

A kettle of heat capacity 400Jkg 'C' contains 2kg of water at a temperature of 30°C. The great heat capacity of water is 4000Jkg °C'. A 2kW heating coil was used to raise the water temperature to 80 °C. If there is no thermal damage to the environment, find the time taken by the coil for this process.

Learn how to calculate the time required by a 2kW heating coil to heat 2kg of water from 30°C to 80°C. This solution covers thermodynamic concepts, heat transfer, and energy calculations, suitable for high school students studying Grades 10-12.

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