A calorimeter of negligible heat capacity contains 100cc of water at `40^(@)C` . The water cools to `35^(@)C` in 5minutes. The water is now replaced by k-oil of equal volume at `40^(@)C` . Find the time taken for the temperature to become `35^(@)C` under similar conditions. Specific heat cpacities of water and K-oil are `4200Jkg^(-1)K^(-1)` and `2100Jkg^(-1)K^(-1)` respectively. Density of K-oil `=800kgm^(-3)` .

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the mass of water
The volume of water is given as 100 cc (which is equivalent to 100 mL or 100 x 10^(-6) m³). The density of water is approximately 1000 kg/m³.

\[ \text{Mass of water (m)} = \text{Density} \times \text{Volume} = 1000 \, \text{kg/m}^3 \times 100 \times 10^{-6} \, \text{m}^3 = 0.1 \, \text{kg} \]

### Step 2: Calculate the heat lost by water
The specific heat capacity of water (c_water) is given as 4200 J/(kg·K). The change in temperature (ΔT) for water is from 40°C to 35°C, which is a change of 5°C or 5 K.

\[ \text{Heat lost by water (Q)} = m \cdot c_{water} \cdot \Delta T = 0.1 \, \text{kg} \times 4200 \, \text{J/(kg·K)} \times 5 \, \text{K} = 2100 \, \text{J} \]

### Step 3: Determine the time taken for water to cool
The time taken for the water to cool from 40°C to 35°C is given as 5 minutes (300 seconds).

### Step 4: Set up the equation for K-oil
Now, we replace water with K-oil of the same volume (100 cc). The density of K-oil is given as 800 kg/m³, and its specific heat capacity (c_oil) is 2100 J/(kg·K).

### Step 5: Determine the mass of K-oil
\[ \text{Mass of K-oil (m_oil)} = \text{Density} \times \text{Volume} = 800 \, \text{kg/m}^3 \times 100 \times 10^{-6} \, \text{m}^3 = 0.08 \, \text{kg} \]

### Step 6: Calculate the heat lost by K-oil
The change in temperature (ΔT) for K-oil is also from 40°C to 35°C, which is again 5 K.

\[ \text{Heat lost by K-oil (Q_oil)} = m_{oil} \cdot c_{oil} \cdot \Delta T = 0.08 \, \text{kg} \times 2100 \, \text{J/(kg·K)} \times 5 \, \text{K} = 840 \, \text{J} \]

### Step 7: Relate the heat lost to time
From the previous calculation, we know that the heat lost by water in 300 seconds is 2100 J. We can set up a ratio to find the time taken for K-oil to lose 840 J.

\[ \frac{Q_{water}}{t_{water}} = \frac{Q_{oil}}{t_{oil}} \]
\[ \frac{2100 \, \text{J}}{300 \, \text{s}} = \frac{840 \, \text{J}}{t_{oil}} \]

### Step 8: Solve for \( t_{oil} \)
Cross-multiplying gives:

\[ 2100 \, \text{J} \cdot t_{oil} = 840 \, \text{J} \cdot 300 \, \text{s} \]
\[ t_{oil} = \frac{840 \times 300}{2100} = 120 \, \text{s} \]

### Step 9: Convert time to minutes
\[ t_{oil} = \frac{120 \, \text{s}}{60} = 2 \, \text{minutes} \]

### Final Answer
The time taken for the K-oil to cool from 40°C to 35°C is **2 minutes**.

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