A ball of thremal capacity `10 cal//^@C` is heated to the temperature of furnace it is then transferred into a vessel containing water. The water equivalent of vessel and the constents is 200 g. The temperature of the vessel and its contents rises from `10^@C` to `40^@C`. What is the temperature of furnace?

Options

A. `640^@C`
B. `64^@C`
C. `600^@C`
D. `100^@C`

Explanation

To solve the problem step by step, we will use the principle of calorimetry, which states that the heat lost by the hot body (the ball) is equal to the heat gained by the cold body (the water in the vessel).

### Step 1: Understand the given data
- Thermal capacity of the ball (C_ball) = 10 cal/°C
- Water equivalent of the vessel and contents (W) = 200 g
- Initial temperature of the vessel and contents (T_initial) = 10°C
- Final temperature of the vessel and contents (T_final) = 40°C
- Let the temperature of the furnace be T_f.

### Step 2: Calculate the heat gained by the water
The heat gained by the water can be calculated using the formula:
\[ Q_{gained} = W \cdot (T_{final} - T_{initial}) \]
Substituting the values:
\[ Q_{gained} = 200 \, \text{g} \cdot (40°C - 10°C) = 200 \cdot 30 = 6000 \, \text{cal} \]

### Step 3: Calculate the heat lost by the ball
The heat lost by the ball can be calculated using the formula:
\[ Q_{lost} = C_{ball} \cdot (T_f - T_{final}) \]
Substituting the values:
\[ Q_{lost} = 10 \cdot (T_f - 40) \]

### Step 4: Set up the equation using the principle of calorimetry
According to the principle of calorimetry:
\[ Q_{lost} = Q_{gained} \]
Thus, we can set up the equation:
\[ 10 \cdot (T_f - 40) = 6000 \]

### Step 5: Solve for T_f
Now, we will solve for T_f:
\[ 10(T_f - 40) = 6000 \]
Dividing both sides by 10:
\[ T_f - 40 = 600 \]
Adding 40 to both sides:
\[ T_f = 640°C \]

### Conclusion
The temperature of the furnace is **640°C**.