The amount of heat conducted out per second through a window, when inside temperature is `10^(@)C` and outside temperature is `-10^(@)C`, is 1000 J. Same heat will be conducted in through the window, when outside temperature is `-23^(@)C` and inside temperature is:

Options

A. `23^(@)C`
B. `230K`
C. `270K`
D. `296K`

Explanation

To solve the problem, we will use the principle of heat conduction through a window, which can be expressed using the formula:

\[ \frac{Q}{t} = kA \frac{T_1 - T_0}{L} \]

Where:
- \( \frac{Q}{t} \) is the rate of heat transfer (in Joules per second),
- \( k \) is the thermal conductivity,
- \( A \) is the area of the window,
- \( T_1 \) is the inside temperature,
- \( T_0 \) is the outside temperature,
- \( L \) is the thickness of the window.

### Step 1: Identify the known values
From the problem statement, we have:
- Inside temperature \( T_1 = 10^\circ C = 283 \, K \)
- Outside temperature \( T_0 = -10^\circ C = 263 \, K \)
- Rate of heat transfer \( \frac{Q}{t} = 1000 \, J/s \)

### Step 2: Write the equation for the first case
Using the formula for heat conduction:
\[ 1000 = kA \frac{283 - 263}{L} \]
This simplifies to:
\[ 1000 = kA \frac{20}{L} \]

### Step 3: Set up the equation for the second case
In the second case, we have:
- New outside temperature \( T_0' = -23^\circ C = 250 \, K \)
- We need to find the new inside temperature \( T_2 \).

Using the same heat conduction formula:
\[ \frac{Q}{t} = kA \frac{T_2 - T_0'}{L} \]
This gives us:
\[ 1000 = kA \frac{T_2 - 250}{L} \]

### Step 4: Equate the two cases
Since the rate of heat transfer is the same in both cases, we can set the two equations equal to each other:
\[ kA \frac{20}{L} = kA \frac{T_2 - 250}{L} \]

### Step 5: Cancel out common terms
We can cancel \( kA \) and \( L \) from both sides (assuming they are constant):
\[ 20 = T_2 - 250 \]

### Step 6: Solve for \( T_2 \)
Rearranging the equation gives:
\[ T_2 = 20 + 250 = 270 \, K \]

### Step 7: Convert \( T_2 \) back to Celsius
To convert back to Celsius:
\[ T_2 = 270 \, K - 273 = -3^\circ C \]

### Final Answer
The inside temperature when the outside temperature is \(-23^\circ C\) will be \(-3^\circ C\).