Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of 0.1 gm/sec. It melts completely in 100 sec. The rate of rise of temperature thereafter will be (Assume no loss of heat)

Options

A. `0.8 .°C//sec`
B. `5.4 .°C//sec`
C. `3.6 .°C//sec`
D. will change with time

Answer

To solve the problem, we need to determine the rate of rise of temperature of the water after the ice has completely melted. Here’s a step-by-step solution:

### Step 1: Calculate the total mass of ice melted
The problem states that the ice melts at a rate of 0.1 grams per second for 100 seconds.

\[ \text{Total mass of ice melted} = \text{melting rate} \times \text{time} = 0.1 \, \text{g/s} \times 100 \, \text{s} = 10 \, \text{g} \]

### Step 2: Determine the heat required to melt the ice
The heat required to melt ice can be calculated using the formula:

\[ Q = m \times L_f \]

where:
- \( Q \) is the heat required,
- \( m \) is the mass of ice (10 g),
- \( L_f \) is the latent heat of fusion of ice, which is approximately \( 334 \, \text{J/g} \).

Calculating the heat required:

\[ Q = 10 \, \text{g} \times 334 \, \text{J/g} = 3340 \, \text{J} \]

### Step 3: Determine the rate of heat supplied
Since the heat is supplied at a constant rate, we can find the rate of heat supplied (\( \frac{dQ}{dt} \)):

\[ \frac{dQ}{dt} = \frac{Q}{\text{time}} = \frac{3340 \, \text{J}}{100 \, \text{s}} = 33.4 \, \text{J/s} \]

### Step 4: Calculate the rate of rise of temperature after melting
After the ice has melted, the heat supplied will now raise the temperature of the resulting water. We can use the formula:

\[ \frac{dQ}{dt} = m \times c \times \frac{dT}{dt} \]

where:
- \( c \) is the specific heat capacity of water, approximately \( 4.18 \, \text{J/g°C} \),
- \( m \) is the mass of water (10 g).

Rearranging the formula to find \( \frac{dT}{dt} \):

\[ \frac{dT}{dt} = \frac{\frac{dQ}{dt}}{m \times c} \]

Substituting the known values:

\[ \frac{dT}{dt} = \frac{33.4 \, \text{J/s}}{10 \, \text{g} \times 4.18 \, \text{J/g°C}} = \frac{33.4}{41.8} \approx 0.8 \, \text{°C/s} \]

### Final Answer
The rate of rise of temperature thereafter will be approximately \( 0.8 \, \text{°C/s} \).

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