A metal ball of mass 1 kg is heated by means of a 20 W heater in a room at `20^(@)C`. The temperature of the ball becomes steady at `50^(@)C`. Assume newton's law of cooling to hold good in the given situation. The temperature of the ball rises uniformly from `20^(@)C` to `30^(@)C` in 5 minutes. select the correct alternatives
(1) The rate of heat loss by ball to surrounding is 20 W, when it is at `50^(@)`
(2) The rate of heat loss by ball to surrounding is `(20)/(3)W`, when it is at `30^(@)C`
(3) The rate of heat loss by ball to surrounding 20 W, when it is at `30^(@)C`
(4). The specific heat capacity of the gas is 500 J/kg K.

Answer

To solve the problem step by step, we will analyze the information given and apply Newton's law of cooling.

### Step 1: Understand the Problem
We have a metal ball of mass 1 kg heated by a 20 W heater. The room temperature is 20°C, and the ball reaches a steady temperature of 50°C. The ball's temperature increases from 20°C to 30°C in 5 minutes. We need to determine the rate of heat loss at two temperatures (30°C and 50°C) and the specific heat capacity of the gas.

### Step 2: Apply Newton's Law of Cooling
Newton's law of cooling states that the rate of heat loss (dQ/dt) is proportional to the temperature difference between the object and its surroundings.

The formula is:
\[ \frac{dQ}{dt} = k \cdot (T - T_s) \]
where:
- \( T \) is the temperature of the ball,
- \( T_s \) is the surrounding temperature (20°C),
- \( k \) is a constant.

### Step 3: Find the Constant \( k \)
When the ball is at 50°C:
- \( T = 50°C \)
- \( T_s = 20°C \)

Using the heater power:
\[ \frac{dQ}{dt} = 20 \, \text{W} \]
Thus,
\[ 20 = k \cdot (50 - 20) \]
\[ 20 = k \cdot 30 \]
\[ k = \frac{20}{30} = \frac{2}{3} \]

### Step 4: Calculate Heat Loss at 30°C
Now, we need to find the rate of heat loss when the ball is at 30°C.
Using the same formula:
\[ \frac{dQ}{dt} = k \cdot (T - T_s) \]
Substituting \( T = 30°C \):
\[ \frac{dQ}{dt} = \frac{2}{3} \cdot (30 - 20) \]
\[ \frac{dQ}{dt} = \frac{2}{3} \cdot 10 = \frac{20}{3} \, \text{W} \]

### Step 5: Calculate Heat Loss at 50°C
We already calculated the heat loss at 50°C:
\[ \frac{dQ}{dt} = k \cdot (50 - 20) \]
\[ \frac{dQ}{dt} = \frac{2}{3} \cdot 30 = 20 \, \text{W} \]

### Step 6: Calculate Specific Heat Capacity
The ball's temperature rises from 20°C to 30°C in 5 minutes (300 seconds). The average rate of heat absorbed can be calculated as:
\[ \text{Average } \frac{dQ}{dt} = \frac{0 + \frac{20}{3}}{2} = \frac{10}{3} \, \text{W} \]

Total heat absorbed in 5 minutes:
\[ Q = \text{Average } \frac{dQ}{dt} \times \text{time} = \frac{10}{3} \times 300 = 1000 \, \text{J} \]

Using the heat balance:
\[ Q_{\text{absorbed}} = Q_{\text{supplied}} - Q_{\text{lost}} \]
\[ 1000 = 20 \times 300 - Q_{\text{lost}} \]
\[ 1000 = 6000 - Q_{\text{lost}} \]
\[ Q_{\text{lost}} = 6000 - 1000 = 5000 \, \text{J} \]

Using the formula for heat absorbed:
\[ Q = m \cdot C \cdot \Delta T \]
Where \( \Delta T = 30 - 20 = 10 \, \text{K} \):
\[ 5000 = 1 \cdot C \cdot 10 \]
\[ C = \frac{5000}{10} = 500 \, \text{J/kg K} \]

### Conclusion
1. The rate of heat loss by the ball to surrounding is 20 W when it is at 50°C. (True)
2. The rate of heat loss by the ball to surrounding is \( \frac{20}{3} \) W when it is at 30°C. (True)
3. The rate of heat loss by the ball to surrounding is 20 W when it is at 30°C. (False)
4. The specific heat capacity of the gas is 500 J/kg K. (True)