A person weighing 50 kg takes in 1500 kcal dict per day. If this energy were to be used in heating the body of person without any losses, then the rise in his temperature is ( specific heat of human body `= 0.83 calg^(-1)C^(-1)`)

Options

A. `30 .^(@)C`
B. `48 .^(@)C`
C. `40.16.^(@)C`
D. `36.14 .^(@)C`

Explanation

To solve the problem of determining the rise in temperature of a person weighing 50 kg who takes in 1500 kcal of energy per day, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the mass from kg to grams**:
\[ \text{Mass (m)} = 50 \, \text{kg} = 50 \times 10^3 \, \text{g} = 50000 \, \text{g} \]

2. **Convert the energy intake from kcal to calories**:
\[ \text{Energy intake} = 1500 \, \text{kcal} = 1500 \times 10^3 \, \text{cal} = 1500000 \, \text{cal} \]

3. **Use the formula for heat transfer**:
The formula for heat transfer is given by:
\[ \Delta Q = m \cdot s \cdot \Delta T \]
where:
- \(\Delta Q\) is the heat energy (in calories),
- \(m\) is the mass (in grams),
- \(s\) is the specific heat capacity (in cal/g°C),
- \(\Delta T\) is the change in temperature (in °C).

4. **Rearrange the formula to solve for \(\Delta T\)**:
\[ \Delta T = \frac{\Delta Q}{m \cdot s} \]

5. **Substitute the known values into the equation**:
- \(\Delta Q = 1500000 \, \text{cal}\)
- \(m = 50000 \, \text{g}\)
- \(s = 0.83 \, \text{cal/g°C}\)

Thus,
\[ \Delta T = \frac{1500000}{50000 \times 0.83} \]

6. **Calculate \(\Delta T\)**:
\[ \Delta T = \frac{1500000}{41500} \approx 36.14 \, \text{°C} \]

### Final Answer:
The rise in temperature of the person is approximately **36.14 °C**.

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