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 calculating the heat required to raise the temperature of 2 g of a substance from 5°C to 15°C, we will follow these steps:

### Step 1: Understand the formula for heat
The heat \( Q \) required to change the temperature of a substance can be calculated using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( m \) is the mass of the substance (in grams),
- \( c \) is the specific heat capacity (which varies with temperature in this case),
- \( \Delta T \) is the change in temperature (in °C).

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

### Step 3: Set up the integral for heat calculation
Since the specific heat varies with temperature, we need to integrate the specific heat over the temperature range from \( 5°C \) to \( 15°C \):
\[ Q = m \int_{t_1}^{t_2} c(t) \, dt \]
where \( t_1 = 5°C \) and \( t_2 = 15°C \).

### Step 4: Substitute the specific heat function into the integral
Substituting the specific heat function into the integral:
\[ Q = 2 \int_{5}^{15} (0.20 + 0.14t + 0.023t^2) \, dt \]

### Step 5: Calculate the integral
Now we calculate the integral:
\[ \int (0.20 + 0.14t + 0.023t^2) \, dt = 0.20t + 0.07t^2 + \frac{0.023}{3}t^3 \]
Now, we evaluate this from \( 5 \) to \( 15 \):
\[ Q = 2 \left[ \left(0.20(15) + 0.07(15^2) + \frac{0.023}{3}(15^3)\right) - \left(0.20(5) + 0.07(5^2) + \frac{0.023}{3}(5^3)\right) \right] \]

### Step 6: Calculate the values at the limits
Calculating the upper limit (at \( t = 15 \)):
\[ = 0.20(15) + 0.07(225) + \frac{0.023}{3}(3375) \]
\[ = 3 + 15.75 + 25.3125 = 44.0625 \]

Calculating the lower limit (at \( t = 5 \)):
\[ = 0.20(5) + 0.07(25) + \frac{0.023}{3}(125) \]
\[ = 1 + 1.75 + 0.95833 = 3.70833 \]

### Step 7: Calculate the heat \( Q \)
Now substituting back into our equation for \( Q \):
\[ Q = 2 \left( 44.0625 - 3.70833 \right) \]
\[ = 2 \times 40.35417 = 80.70834 \text{ calories} \]

### Step 8: Round off the answer
Rounding off gives us approximately \( 81 \) calories.

### Final Answer
The heat required to raise the temperature of 2 g of the substance from 5°C to 15°C is approximately **81 calories**.

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