A person weighing 50 kg takes in 1500 kcal diet 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 step by step, we will use the formula relating heat energy, mass, specific heat, and temperature change. The formula is:

\[ \Delta Q = m \cdot s \cdot \Delta T \]

Where:
- \(\Delta Q\) is the heat energy absorbed (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).

### Step 1: Convert the mass from kg to grams
The mass of the person is given as 50 kg. To convert this to grams, we use the conversion factor \(1 \text{ kg} = 1000 \text{ g}\):

\[ m = 50 \, \text{kg} \times 1000 \, \text{g/kg} = 50000 \, \text{g} \]

### Step 2: Identify the energy intake in calories
The energy intake is given as 1500 kcal per day. We need to convert this into calories since the specific heat is given in cal/g°C. The conversion is:

\[ 1 \text{ kcal} = 1000 \text{ cal} \]

Thus,

\[ \Delta Q = 1500 \, \text{kcal} \times 1000 \, \text{cal/kcal} = 1500000 \, \text{cal} \]

### Step 3: Use the specific heat value
The specific heat of the human body is given as \(s = 0.83 \, \text{cal/g°C}\).

### Step 4: Rearrange the formula to find \(\Delta T\)
From the formula \(\Delta Q = m \cdot s \cdot \Delta T\), we can rearrange it to find \(\Delta T\):

\[ \Delta T = \frac{\Delta Q}{m \cdot s} \]

### Step 5: Substitute the values into the equation
Now, substituting the values we have:

\[ \Delta T = \frac{1500000 \, \text{cal}}{50000 \, \text{g} \cdot 0.83 \, \text{cal/g°C}} \]

### Step 6: Calculate \(\Delta T\)
Calculating the denominator:

\[ 50000 \, \text{g} \cdot 0.83 \, \text{cal/g°C} = 41500 \, \text{cal/°C} \]

Now substituting back into the equation for \(\Delta T\):

\[ \Delta T = \frac{1500000 \, \text{cal}}{41500 \, \text{cal/°C}} \approx 36.14 \, °C \]

### Final Answer
The rise in temperature of the person is approximately \(36.14 \, °C\).

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