A certain substance has a mass per mole of 50.0 `g//` mol. When 325 J is added as heat to a 30.0 g sample, the sample's temperature rises from `25.0^(@)` to `45.0^(@)` C. What are the (a) specific heat and (b) molar specific heat of this substance ? (c ) How many moles are in the sample ?

Answer

To solve the problem step by step, we will find the specific heat, molar specific heat, and the number of moles in the sample.

### Given Data:
- Mass per mole (molar mass) = 50.0 g/mol
- Mass of the sample = 30.0 g
- Heat added (Q) = 325 J
- Initial temperature (T1) = 25.0 °C
- Final temperature (T2) = 45.0 °C

### Step 1: Calculate the change in temperature (ΔT)
\[ \Delta T = T_2 - T_1 = 45.0 °C - 25.0 °C = 20.0 °C \]

### Step 2: Calculate the specific heat (c)
The formula for specific heat is given by:
\[ c = \frac{Q}{m \cdot \Delta T} \]
Where:
- \( Q \) = heat added (325 J)
- \( m \) = mass of the sample (30.0 g = 0.030 kg)
- \( \Delta T \) = change in temperature (20.0 °C)

Substituting the values:
\[ c = \frac{325 \, \text{J}}{0.030 \, \text{kg} \cdot 20.0 \, \text{°C}} = \frac{325}{0.600} \approx 541.67 \, \text{J/(kg·K)} \]

### Step 3: Calculate the number of moles (n)
The number of moles can be calculated using the formula:
\[ n = \frac{m}{\text{molar mass}} \]
Substituting the values:
\[ n = \frac{30.0 \, \text{g}}{50.0 \, \text{g/mol}} = 0.6 \, \text{mol} \]

### Step 4: Calculate the molar specific heat (C)
The formula for molar specific heat is given by:
\[ C = \frac{Q}{n \cdot \Delta T} \]
Where:
- \( n \) = number of moles (0.6 mol)

Substituting the values:
\[ C = \frac{325 \, \text{J}}{0.6 \, \text{mol} \cdot 20.0 \, \text{°C}} = \frac{325}{12} \approx 27.08 \, \text{J/(mol·K)} \]

### Final Answers:
(a) Specific heat \( c \approx 541.67 \, \text{J/(kg·K)} \)

(b) Molar specific heat \( C \approx 27.08 \, \text{J/(mol·K)} \)

(c) Number of moles \( n = 0.6 \, \text{mol} \)

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