Ram ans Shyam purchased two electric tea kettles A and B of same size, same thickness, and same volume of 0.4L. They studied the specification of kettles as under
`Kettle A: `
specific heat capacity `= 1680 Jkg^(-1)K^(-1)`
Mass = 200 g
Cost = Rs. 400
Kettle B:
Specific heat capacity = `2450 Jkg^(-1) K^(-1)`
Mass = 400 g
Cost = Rs. 400
When kettle A is switched on with constant potential source, the tea begins to boil in 6 min. When kettle B is switched on with the same source separately, then tea begins to boil in 8 min. The efficiengy of kettle is defined as
(Energy used for liquid heating)/(Total energy supplied)
They made discussion on specification and efficiency of kettles and subsequently prepared a list of questions to draw the conclusions. Some of them are as under (assume specific heat of tea liquid as `4200J kg^(-1)K^(-1)` and density `1000 kgm^(-3)`
Efficiency of kettle A is

Options

A. `63.34%`
B. `83.34%`
C. `93.34%`
D. `73.34%`

Explanation

To find the efficiency of kettle A, we will follow these steps:

### Step 1: Calculate the mass of water in kettle A
Given that the volume of kettle A is 0.4 L, we can convert this to cubic meters:
\[ \text{Volume} = 0.4 \, \text{L} = 0.4 \times 10^{-3} \, \text{m}^3 \]
Using the density of water (1000 kg/m³), we can find the mass of the water:
\[ \text{Mass of water} (M_w) = \text{Density} \times \text{Volume} = 1000 \, \text{kg/m}^3 \times 0.4 \times 10^{-3} \, \text{m}^3 = 0.4 \, \text{kg} = 400 \, \text{g} \]

### Step 2: Calculate the heat absorbed by the water
The specific heat capacity of water is given as 4200 J/(kg·K). The heat absorbed by the water (Q_w) can be calculated using the formula:
\[ Q_w = M_w \cdot c_w \cdot \Delta T \]
Assuming the change in temperature (\(\Delta T\)) is \(T\):
\[ Q_w = 0.4 \, \text{kg} \cdot 4200 \, \text{J/(kg·K)} \cdot T = 1680 \, \text{J/K} \cdot T \]

### Step 3: Calculate the heat absorbed by the kettle
The mass of kettle A is given as 200 g (0.2 kg) and its specific heat capacity is 1680 J/(kg·K). The heat absorbed by the kettle (Q_k) is:
\[ Q_k = M_k \cdot c_k \cdot \Delta T \]
Where:
- \(M_k = 0.2 \, \text{kg}\)
- \(c_k = 1680 \, \text{J/(kg·K)}\)
- \(\Delta T = T\)

Thus,
\[ Q_k = 0.2 \, \text{kg} \cdot 1680 \, \text{J/(kg·K)} \cdot T = 336 \, \text{J/K} \cdot T \]

### Step 4: Calculate the total heat supplied
The total heat supplied (Q_total) is the sum of the heat absorbed by the kettle and the water:
\[ Q_{\text{total}} = Q_k + Q_w = 336 \, \text{J/K} \cdot T + 1680 \, \text{J/K} \cdot T = (336 + 1680) \, \text{J/K} \cdot T = 2016 \, \text{J/K} \cdot T \]

### Step 5: Calculate the efficiency
The efficiency (\(\eta\)) is defined as the ratio of the heat used for heating the liquid to the total energy supplied:
\[ \eta = \frac{Q_w}{Q_{\text{total}}} \]
Substituting the values we calculated:
\[ \eta = \frac{1680 \, \text{J/K} \cdot T}{2016 \, \text{J/K} \cdot T} \]
The \(T\) cancels out:
\[ \eta = \frac{1680}{2016} \]
Calculating this gives:
\[ \eta \approx 0.8333 \text{ or } 83.33\% \]

### Conclusion
Thus, the efficiency of kettle A is approximately **83.34%**.

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