When a `290 gm` piece of iron at `180^(@)C` is placed in a `100 gm` aluminium calorimeter cup containing `250 gm` of glycerin at `10^(@)C`, the final temperature is observed to be `38^(@)C`. What is the specific heat of glycerin? Use specific heat of iron 470 J/k K and that of aluminium is 900 J/kg K.

Answer

To solve the problem, we will apply the principle of conservation of energy, which states that the total heat lost by the hot object (iron) will be equal to the total heat gained by the cold objects (aluminium calorimeter and glycerin).

### Step 1: Identify the known values
- Mass of iron, \( m_{Fe} = 290 \, \text{g} = 0.290 \, \text{kg} \)
- Initial temperature of iron, \( T_{i,Fe} = 180^\circ C \)
- Final temperature, \( T_f = 38^\circ C \)
- Specific heat of iron, \( c_{Fe} = 470 \, \text{J/kg K} \)

- Mass of aluminium, \( m_{Al} = 100 \, \text{g} = 0.100 \, \text{kg} \)
- Initial temperature of aluminium, \( T_{i,Al} = 10^\circ C \)
- Specific heat of aluminium, \( c_{Al} = 900 \, \text{J/kg K} \)

- Mass of glycerin, \( m_{G} = 250 \, \text{g} = 0.250 \, \text{kg} \)
- Initial temperature of glycerin, \( T_{i,G} = 10^\circ C \)
- Specific heat of glycerin, \( c_G = ? \) (unknown)

### Step 2: Write the heat gained and lost equations
The heat lost by the iron can be calculated as:
\[ Q_{lost,Fe} = m_{Fe} \cdot c_{Fe} \cdot (T_{i,Fe} - T_f) \]
The heat gained by the aluminium and glycerin can be calculated as:
\[ Q_{gained,Al} = m_{Al} \cdot c_{Al} \cdot (T_f - T_{i,Al}) \]
\[ Q_{gained,G} = m_{G} \cdot c_G \cdot (T_f - T_{i,G}) \]

### Step 3: Set up the energy conservation equation
According to the principle of conservation of energy:
\[ Q_{lost,Fe} = Q_{gained,Al} + Q_{gained,G} \]

Substituting the equations from Step 2:
\[ m_{Fe} \cdot c_{Fe} \cdot (T_{i,Fe} - T_f) = m_{Al} \cdot c_{Al} \cdot (T_f - T_{i,Al}) + m_{G} \cdot c_G \cdot (T_f - T_{i,G}) \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ 0.290 \cdot 470 \cdot (180 - 38) = 0.100 \cdot 900 \cdot (38 - 10) + 0.250 \cdot c_G \cdot (38 - 10) \]

### Step 5: Calculate the left side
Calculating the left side:
\[ 0.290 \cdot 470 \cdot 142 = 0.290 \cdot 470 \cdot 142 = 24050.6 \, \text{J} \]

### Step 6: Calculate the right side
Calculating the right side:
\[ 0.100 \cdot 900 \cdot 28 = 2520 \, \text{J} \]
\[ 0.250 \cdot c_G \cdot 28 = 7c_G \, \text{J} \]

### Step 7: Set up the equation
Now we can set up the equation:
\[ 24050.6 = 2520 + 7c_G \]

### Step 8: Solve for \( c_G \)
Rearranging the equation to solve for \( c_G \):
\[ 24050.6 - 2520 = 7c_G \]
\[ 21530.6 = 7c_G \]
\[ c_G = \frac{21530.6}{7} \approx 3061.5 \, \text{J/kg K} \]

### Final Answer
The specific heat of glycerin is approximately \( 3061.5 \, \text{J/kg K} \).

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