Questions: Calculate energy required to heat 201.0 g of iron from -3.2°C to 6.8°C. Assume the specific heat capacity of iron under these conditions is 0.449 J · g.

Calculate energy required to heat 201.0 g of iron from -3.2°C to 6.8°C. Assume the specific heat capacity of iron under these conditions is 0.449 J · g.

Transcript text: Calculate energy required to heat 201.0 g of iron from $-3.2^{\circ} \mathrm{C}$ to $6.8^{\circ} \mathrm{C}$. Assume the specific heat capacity of iron under these conditions is $0.449 \mathrm{~J} \cdot \mathrm{~g}$.

Solution

Solution Steps

Step 1: Identify the given values

We are given:

Mass of iron, $ m = 201.0 $ g
Initial temperature, $ T_i = -3.2^{\circ} \mathrm{C} $
Final temperature, $ T_f = 6.8^{\circ} \mathrm{C} $
Specific heat capacity of iron, $ c = 0.449 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~^{\circ}C}^{-1} $

Step 2: Calculate the temperature change

The temperature change, $ \Delta T $, is given by: \[ \Delta T = T_f - T_i = 6.8^{\circ} \mathrm{C} - (-3.2^{\circ} \mathrm{C}) = 6.8 + 3.2 = 10.0^{\circ} \mathrm{C} \]

Step 3: Use the formula for heat energy

The formula to calculate the energy required to heat the iron is: \[ Q = mc\Delta T \] where:

$ Q $ is the heat energy
$ m $ is the mass
$ c $ is the specific heat capacity
$ \Delta T $ is the temperature change

Step 4: Substitute the values into the formula

Substitute the given values into the formula: \[ Q = (201.0 \, \mathrm{g}) \times (0.449 \, \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~^{\circ}C}^{-1}) \times (10.0^{\circ} \mathrm{C}) \]

Step 5: Perform the calculation

\[ Q = 201.0 \times 0.449 \times 10.0 = 902.49 \, \mathrm{J} \]