The temperature of a room heated by heater is `20^@C` when outside temperature is `-20^@C` and it is `10^@C` when the outside temperature is `-40^@C`. The temperature of the heater is

Options

A. `80^@C`
B. `100^@C`
C. `40^@C`
D. `60^@C`

Explanation

To solve the problem, we need to find the temperature of the heater (T) based on the given conditions. We have two scenarios with the corresponding inside and outside temperatures.

### Step 1: Set up the equations based on the given conditions.

1. **Condition 1**: When the outside temperature (T_out1) is -20°C, the inside temperature (T_in1) is 20°C.
2. **Condition 2**: When the outside temperature (T_out2) is -40°C, the inside temperature (T_in2) is 10°C.

We can express the heat balance for both conditions. Assuming that the heat transfer is proportional to the temperature difference, we can write:

For Condition 1:
\[ T - T_{in1} = \alpha (T_{out1} - T) \]
Substituting the values:
\[ T - 20 = \alpha (-20 - T) \]
(1)

For Condition 2:
\[ T - T_{in2} = \alpha (T_{out2} - T) \]
Substituting the values:
\[ T - 10 = \alpha (-40 - T) \]
(2)

### Step 2: Rearrange the equations.

From equation (1):
\[ T - 20 = -\alpha T - 20\alpha \]
\[ T + \alpha T = 20 - 20\alpha \]
\[ T(1 + \alpha) = 20 - 20\alpha \]
\[ T = \frac{20 - 20\alpha}{1 + \alpha} \]
(3)

From equation (2):
\[ T - 10 = -\alpha T - 40\alpha \]
\[ T + \alpha T = 10 - 40\alpha \]
\[ T(1 + \alpha) = 10 - 40\alpha \]
\[ T = \frac{10 - 40\alpha}{1 + \alpha} \]
(4)

### Step 3: Set equations (3) and (4) equal to each other.

Since both equations represent T, we can set them equal:
\[ \frac{20 - 20\alpha}{1 + \alpha} = \frac{10 - 40\alpha}{1 + \alpha} \]

### Step 4: Solve for α.

Cross-multiply:
\[ (20 - 20\alpha)(1 + \alpha) = (10 - 40\alpha)(1 + \alpha) \]

Expanding both sides:
\[ 20 + 20\alpha - 20\alpha - 20\alpha^2 = 10 + 10\alpha - 40\alpha - 40\alpha^2 \]
\[ 20 - 20\alpha^2 = 10 - 30\alpha - 40\alpha^2 \]

Rearranging gives:
\[ 20 + 30\alpha - 10 = 20\alpha^2 \]
\[ 10 + 30\alpha = 20\alpha^2 \]

### Step 5: Rearranging to form a quadratic equation.

Rearranging gives:
\[ 20\alpha^2 - 30\alpha - 10 = 0 \]

### Step 6: Solve the quadratic equation using the quadratic formula.

Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 20, b = -30, c = -10 \):
\[ \alpha = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 20 \cdot (-10)}}{2 \cdot 20} \]
\[ \alpha = \frac{30 \pm \sqrt{900 + 800}}{40} \]
\[ \alpha = \frac{30 \pm \sqrt{1700}}{40} \]
\[ \alpha = \frac{30 \pm 41.23}{40} \]

Calculating the two possible values for α:
1. \( \alpha = \frac{71.23}{40} \) (positive value)
2. \( \alpha = \frac{-11.23}{40} \) (negative value, not valid)

### Step 7: Substitute α back to find T.

Using the positive value of α in either equation (3) or (4) to find T.

### Final Answer:
The temperature of the heater (T) can be calculated from the valid α value.