The specific heat of a substance varies with temperature according to `c =0.2 +0.16 T + 0.024 T^(2)` with `T` in `"^(@)c` and `c` is `cal//gk`. Find the energy (in cal) required to raise the temp of `2g` substance from `0^(@)to 5^(@)C`

Answer

To find the energy required to raise the temperature of a 2g substance from 0°C to 5°C, we will use the specific heat capacity equation given as:

\[ c = 0.2 + 0.16T + 0.024T^2 \]

where \( T \) is the temperature in °C and \( c \) is in cal/gK.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Mass of the substance, \( m = 2 \, \text{g} \)
- Initial temperature, \( T_1 = 0 \, \text{°C} \)
- Final temperature, \( T_2 = 5 \, \text{°C} \)

2. **Set Up the Heat Equation:**
The heat \( Q \) required to change the temperature can be expressed as:
\[ Q = m \int_{T_1}^{T_2} c(T) \, dT \]
where \( c(T) \) is the specific heat capacity as a function of temperature.

3. **Substitute the Specific Heat Capacity:**
Substitute the expression for \( c(T) \):
\[ Q = m \int_{0}^{5} (0.2 + 0.16T + 0.024T^2) \, dT \]
Now, substituting \( m = 2 \, \text{g} \):
\[ Q = 2 \int_{0}^{5} (0.2 + 0.16T + 0.024T^2) \, dT \]

4. **Integrate the Function:**
We will integrate each term separately:
\[ \int (0.2 + 0.16T + 0.024T^2) \, dT = 0.2T + 0.08T^2 + 0.008T^3 + C \]
Now, we evaluate this integral from 0 to 5:
\[ Q = 2 \left[ \left(0.2(5) + 0.08(5^2) + 0.008(5^3)\right) - \left(0.2(0) + 0.08(0^2) + 0.008(0^3)\right) \right] \]

5. **Calculate the Values:**
- Calculate each term at \( T = 5 \):
- \( 0.2(5) = 1.0 \)
- \( 0.08(5^2) = 0.08(25) = 2.0 \)
- \( 0.008(5^3) = 0.008(125) = 1.0 \)

Therefore,
\[ Q = 2 \left[ 1.0 + 2.0 + 1.0 \right] = 2 \times 4.0 = 8.0 \, \text{cal} \]

6. **Final Result:**
The energy required to raise the temperature of the 2g substance from 0°C to 5°C is:
\[ Q = 8 \, \text{cal} \]