The specific heat of a substance varies with temperature according to the function c = 0.20` +` 0.14 T` + 0.023T^(@)`, with T in `""^(@)C` and c in `Cal//g.K`, Find the energy required to raise the temperature of 1.0g of this substance from `5.0 ""^(@)C` to `15""^(@)C`.

Answer

To find the energy required to raise the temperature of 1.0 g of the substance from 5.0 °C to 15.0 °C, we will use the formula for heat transfer:

\[ Q = \int_{T_1}^{T_2} mc(T) \, dT \]

Where:
- \( Q \) is the heat energy required,
- \( m \) is the mass of the substance (1.0 g),
- \( c(T) \) is the specific heat capacity as a function of temperature,
- \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively.

### Step 1: Write down the specific heat function
The specific heat \( c \) is given by:
\[ c(T) = 0.20 + 0.14T + 0.023T^2 \]

### Step 2: Set up the integral for heat transfer
Substituting the specific heat function into the heat transfer equation:
\[ Q = \int_{5}^{15} (1.0) \cdot (0.20 + 0.14T + 0.023T^2) \, dT \]

### Step 3: Break down the integral
This can be separated into three integrals:
\[ Q = \int_{5}^{15} 0.20 \, dT + \int_{5}^{15} 0.14T \, dT + \int_{5}^{15} 0.023T^2 \, dT \]

### Step 4: Calculate each integral
1. **First Integral**:
\[ \int_{5}^{15} 0.20 \, dT = 0.20 \cdot (15 - 5) = 0.20 \cdot 10 = 2.0 \, \text{Cal} \]

2. **Second Integral**:
\[ \int_{5}^{15} 0.14T \, dT = 0.14 \cdot \left[\frac{T^2}{2}\right]_{5}^{15} = 0.14 \cdot \left(\frac{15^2}{2} - \frac{5^2}{2}\right) \]
\[ = 0.14 \cdot \left(\frac{225}{2} - \frac{25}{2}\right) = 0.14 \cdot \left(\frac{200}{2}\right) = 0.14 \cdot 100 = 14.0 \, \text{Cal} \]

3. **Third Integral**:
\[ \int_{5}^{15} 0.023T^2 \, dT = 0.023 \cdot \left[\frac{T^3}{3}\right]_{5}^{15} = 0.023 \cdot \left(\frac{15^3}{3} - \frac{5^3}{3}\right) \]
\[ = 0.023 \cdot \left(\frac{3375}{3} - \frac{125}{3}\right) = 0.023 \cdot \left(\frac{3250}{3}\right) = 0.023 \cdot 1083.33 \approx 24.92 \, \text{Cal} \]

### Step 5: Sum the results
Now, we can sum the results of the three integrals to find the total heat \( Q \):
\[ Q = 2.0 + 14.0 + 24.92 \approx 40.92 \, \text{Cal} \]

### Final Answer
The energy required to raise the temperature of 1.0 g of the substance from 5.0 °C to 15.0 °C is approximately **40.92 Cal**.

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