The specific heat of a substance varies with temperature `t(.^(@)C)` as
`c=0.20+0.14t+0.023t^2 (cal//g^(@)//C)`
The heat required to raise the temperature of 2 g of substance from `5^@C` to `15^@C` will be

Options

A. 24 cal
B. 56 cal
C. 82 cal
D. 100 cal

Explanation

To solve the problem of finding the heat required to raise the temperature of 2 g of a substance from 5°C to 15°C, given that the specific heat varies with temperature, we can follow these steps:

### Step 1: Understand the formula for heat
The heat \( Q \) required to change the temperature of a substance with variable specific heat \( c(t) \) is given by the integral:
\[ Q = \int_{t_1}^{t_2} mc(t) \, dt \]
where \( m \) is the mass of the substance, \( t_1 \) is the initial temperature, and \( t_2 \) is the final temperature.

### Step 2: Identify the specific heat function
The specific heat of the substance is given as:
\[ c(t) = 0.20 + 0.14t + 0.023t^2 \, \text{(cal/g°C)} \]

### Step 3: Set up the integral
For our case, we have:
- Mass \( m = 2 \, \text{g} \)
- Initial temperature \( t_1 = 5 \, \text{°C} \)
- Final temperature \( t_2 = 15 \, \text{°C} \)

Thus, the heat required can be expressed as:
\[ Q = \int_{5}^{15} 2 \cdot (0.20 + 0.14t + 0.023t^2) \, dt \]

### Step 4: Factor out the constant
Factor out the mass from the integral:
\[ Q = 2 \int_{5}^{15} (0.20 + 0.14t + 0.023t^2) \, dt \]

### Step 5: Integrate the function
Now we will integrate the function:
\[ \int (0.20 + 0.14t + 0.023t^2) \, dt = 0.20t + 0.14 \cdot \frac{t^2}{2} + 0.023 \cdot \frac{t^3}{3} \]

### Step 6: Evaluate the integral from 5 to 15
Now we evaluate the integral from 5 to 15:
\[ Q = 2 \left[ \left(0.20 \cdot 15 + 0.14 \cdot \frac{15^2}{2} + 0.023 \cdot \frac{15^3}{3}\right) - \left(0.20 \cdot 5 + 0.14 \cdot \frac{5^2}{2} + 0.023 \cdot \frac{5^3}{3}\right) \right] \]

### Step 7: Calculate each term
Calculating the terms:
1. For \( t = 15 \):
- \( 0.20 \cdot 15 = 3.0 \)
- \( 0.14 \cdot \frac{15^2}{2} = 0.14 \cdot 112.5 = 15.75 \)
- \( 0.023 \cdot \frac{15^3}{3} = 0.023 \cdot 1125 = 25.875 \)

Total for \( t = 15 \):
\[ 3.0 + 15.75 + 25.875 = 44.625 \]

2. For \( t = 5 \):
- \( 0.20 \cdot 5 = 1.0 \)
- \( 0.14 \cdot \frac{5^2}{2} = 0.14 \cdot 12.5 = 1.75 \)
- \( 0.023 \cdot \frac{5^3}{3} = 0.023 \cdot \frac{125}{3} \approx 0.9583 \)

Total for \( t = 5 \):
\[ 1.0 + 1.75 + 0.9583 \approx 3.7083 \]

### Step 8: Calculate the final heat \( Q \)
Now substituting back:
\[ Q = 2 \left( 44.625 - 3.7083 \right) = 2 \cdot 40.9167 \approx 81.8334 \, \text{cal} \]

### Step 9: Round to the nearest whole number
Thus, rounding gives us:
\[ Q \approx 82 \, \text{cal} \]

### Final Answer
The heat required to raise the temperature of 2 g of the substance from 5°C to 15°C is approximately **82 calories**.

---