An object is being heated by a heater supplying `60W` of heat. Temperature of surrounding is `20^(@)C` and the temperature of object becomes constant at `50^(@)C` Now the heater is switched off What is the rate at which the object will lose heat when its temperature has dropped to `30^(@)C` .

Options

A. ` 20W`
B. `30 W`
C. `40 W`
D. `60 W`

Answer

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

### Step 1: Understand the Initial Conditions
The heater supplies a constant power of 60 W to maintain the temperature of the object at 50°C. The surrounding temperature is 20°C.

### Step 2: Determine the Heat Transfer Coefficient
When the heater is on, the heat loss from the object is equal to the power supplied by the heater. Thus, at 50°C, the rate of heat loss (dQ/dt) is 60 W.

Using Newton's Law of Cooling:
\[ \frac{dQ}{dt} = hA(T - T_0) \]
where:
- \( h \) is the heat transfer coefficient,
- \( A \) is the surface area,
- \( T \) is the temperature of the object,
- \( T_0 \) is the ambient temperature (20°C).

At 50°C:
\[ 60 = hA(50 - 20) \]
\[ 60 = hA(30) \]
Thus, we can express \( hA \):
\[ hA = \frac{60}{30} = 2 \, \text{W/°C} \]

### Step 3: Calculate the Rate of Heat Loss at 30°C
Now, when the heater is switched off, we need to find the rate of heat loss when the temperature of the object drops to 30°C.

Using the same formula:
\[ \frac{dQ}{dt} = hA(T - T_0) \]
Substituting the values:
- \( T = 30°C \)
- \( T_0 = 20°C \)

So:
\[ \frac{dQ}{dt} = hA(30 - 20) \]
\[ \frac{dQ}{dt} = hA(10) \]

Now substituting \( hA = 2 \):
\[ \frac{dQ}{dt} = 2(10) = 20 \, \text{W} \]

### Conclusion
The rate at which the object will lose heat when its temperature has dropped to 30°C is **20 W**.

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