A vessel with 100 g of water at a temperature of `0^(@)C` is suspended in the middle of a room. In 15 minutes the temperature of the water rises to `1.8^(@)C`. When ice equal in weight of the water is placed in the same vessel, it melts during 10 hours. using appropriate appropriate approximations, estimate the latent heat of fusio of ice in cal/g. if the known alue of latent heat of fusion of ice is 80 cal/g, obtain the difference in the two values in cal/g and report this as your answer.

Answer

To solve the problem step by step, we will calculate the heat supplied to the water and then use that to find the latent heat of fusion of ice.

### Step 1: Calculate the heat supplied to the water
The heat supplied to the water can be calculated using the formula:
\[ Q = mc\Delta T \]
Where:
- \( m = 100 \, \text{g} \) (mass of water)
- \( c = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 1.8 \, \text{°C} - 0 \, \text{°C} = 1.8 \, \text{°C} \) (change in temperature)

Substituting the values:
\[ Q = 100 \, \text{g} \times 1 \, \text{cal/g°C} \times 1.8 \, \text{°C} \]
\[ Q = 180 \, \text{cal} \]

### Step 2: Calculate the rate of heat supply
The heat supplied to the water in 15 minutes is 180 cal. To find the rate of heat supply per minute:
\[ \text{Rate of heat supply} = \frac{Q}{\text{time}} = \frac{180 \, \text{cal}}{15 \, \text{min}} = 12 \, \text{cal/min} \]

### Step 3: Calculate the total heat supplied in 10 hours
Since the ice melts in 10 hours, we need to convert hours to minutes:
\[ 10 \, \text{hours} = 10 \times 60 = 600 \, \text{minutes} \]

Now, calculate the total heat supplied in 10 hours:
\[ \text{Total heat} = \text{Rate of heat supply} \times \text{time} \]
\[ \text{Total heat} = 12 \, \text{cal/min} \times 600 \, \text{min} = 7200 \, \text{cal} \]

### Step 4: Relate the heat to the latent heat of fusion of ice
The heat required to melt the ice is given by:
\[ Q = mL \]
Where:
- \( m = 100 \, \text{g} \) (mass of ice)
- \( L \) is the latent heat of fusion of ice.

Setting the two equations equal:
\[ 7200 \, \text{cal} = 100 \, \text{g} \times L \]
Solving for \( L \):
\[ L = \frac{7200 \, \text{cal}}{100 \, \text{g}} = 72 \, \text{cal/g} \]

### Step 5: Calculate the difference from the known value
The known value of the latent heat of fusion of ice is \( 80 \, \text{cal/g} \). The difference is:
\[ \text{Difference} = 80 \, \text{cal/g} - 72 \, \text{cal/g} = 8 \, \text{cal/g} \]

### Final Answer
The difference in the two values of latent heat of fusion of ice is \( 8 \, \text{cal/g} \).

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