A metal block is placed in a room which is at `10^(@)C` for long time. Now it is heated by an electric heater of power 500 W till its temperature becomes `50^(@)C`. Its initial rate of rise of temperature is `2.5^(@)C//sec`. The heater is switched off and now a heater of 100W is required to maintain the temperature of the block at `50^(@)C` . (Assume Newtons Law of cooling to be valid)
What is the rate of cooling of block at `50^(@)C` if the 100W heater is also switched off?

Options

A. `5^(@)C//s`
B. `0.5^(@)C//s`
C. `1^(@)C//s`
D. `0.1^(@)C//s`

Answer

To solve the problem step by step, we will use the principles of heat transfer and Newton's law of cooling.

### Step 1: Understanding the Problem
The metal block is initially at a temperature of 10°C and is heated to 50°C using a 500 W heater. After reaching 50°C, a 100 W heater is needed to maintain this temperature. We need to find the rate of cooling of the block if the 100 W heater is turned off.

### Step 2: Determine the Heat Capacity of the Block
We know that the power of the heater (P) is equal to the rate of change of heat energy (dQ/dt) in the block. According to the problem, the initial rate of rise of temperature is given as 2.5°C/s.

Using the formula:
\[ P = C \frac{dT}{dt} \]
Where:
- \( P = 500 \text{ W} \)
- \( C \) is the heat capacity of the block
- \( \frac{dT}{dt} = 2.5 \text{ °C/s} \)

Substituting the values:
\[ 500 = C \times 2.5 \]

Now, solving for \( C \):
\[ C = \frac{500}{2.5} = 200 \text{ J/°C} \]

### Step 3: Applying Newton's Law of Cooling
When the 100 W heater is used to maintain the temperature at 50°C, we can express the balance of power as:
\[ 0 = C \frac{dT}{dt} + P_{\text{loss}} \]
Where \( P_{\text{loss}} \) is the power lost to the surroundings, which is 100 W when the heater is on.

Substituting the values:
\[ 0 = 200 \frac{dT}{dt} + 100 \]

### Step 4: Solve for the Rate of Cooling
Rearranging the equation to find \( \frac{dT}{dt} \):
\[ 200 \frac{dT}{dt} = -100 \]
\[ \frac{dT}{dt} = -\frac{100}{200} = -0.5 \text{ °C/s} \]

### Conclusion
The rate of cooling of the block at 50°C, when the 100 W heater is switched off, is:
\[ \frac{dT}{dt} = -0.5 \text{ °C/s} \]