If coal gives off 7000 kcal/ kg when it is burnt, how much coal will be needed to heat a house that requires `4.2xx10^(7)` kcal for the whole winter? Assume that an additional 30 percent of the heat is lost up the chimney.

Answer

To solve the problem, we need to calculate the amount of coal required to heat a house that needs `4.2 x 10^7` kcal for the winter, considering that 30% of the heat is lost through the chimney.

### Step-by-Step Solution:

1. **Determine the total heat required considering the heat loss:**
- The house requires `4.2 x 10^7` kcal.
- Since 30% of the heat is lost, we need to find the total heat that must be produced to account for this loss.
- The heat lost is calculated as:
\[ \text{Heat lost} = 0.3 \times (4.2 \times 10^7) = 1.26 \times 10^7 \text{ kcal} \]
- Therefore, the total heat required is:
\[ \text{Total heat required} = \text{Required heat} + \text{Heat lost} = 4.2 \times 10^7 + 1.26 \times 10^7 = 5.46 \times 10^7 \text{ kcal} \]

2. **Calculate the amount of coal needed:**
- We know that 1 kg of coal produces 7000 kcal.
- To find the amount of coal needed, we divide the total heat required by the heat produced per kg of coal:
\[ \text{Amount of coal required} = \frac{\text{Total heat required}}{\text{Heat produced per kg}} = \frac{5.46 \times 10^7}{7000} \]

3. **Perform the division:**
- First, simplify the division:
\[ \frac{5.46 \times 10^7}{7000} = \frac{5.46 \times 10^7}{7 \times 10^3} = \frac{5.46}{7} \times 10^{7-3} = \frac{5.46}{7} \times 10^4 \]
- Now, calculate \( \frac{5.46}{7} \):
\[ \frac{5.46}{7} \approx 0.78 \]
- Therefore, the amount of coal required is:
\[ 0.78 \times 10^4 \text{ kg} = 7.8 \times 10^3 \text{ kg} \]

### Final Answer:
The amount of coal required is approximately **7.8 x 10^3 kg**.