A metal block is placed in a room which is at `10^(@)C` for long time. Now it is heated by an electric heater of power 500 W till its temperature becomes `50^(@)C`. Its initial rate of rise of temperature is `2.5^(@)C//sec`. The heater is switched off and now a heater of 100W is required to maintain the temperature of the block at `50^(@)C` . (Assume Newtons Law of cooling to be valid)
What is the heat capacity of the block?

Options

A. `50J//^(@)C`
B. `100J//^(@)C`
C. `150J//^(@)C`
D. `200J//^(@)C`

Answer

To solve the problem, we need to determine the heat capacity of the metal block based on the information provided. Let's break it down step by step.

### Step 1: Determine the temperature change and the time taken
The initial temperature of the block is \(10^\circ C\) and the final temperature is \(50^\circ C\). The change in temperature (\(\Delta T\)) is:
\[ \Delta T = 50^\circ C - 10^\circ C = 40^\circ C \]

The initial rate of rise of temperature is given as \(2.5^\circ C/s\). To find the time (\(t\)) taken to reach \(50^\circ C\), we can use the formula:
\[ t = \frac{\Delta T}{\text{Rate of rise}} = \frac{40^\circ C}{2.5^\circ C/s} = 16 \text{ seconds} \]

### Step 2: Calculate the heat energy supplied by the heater
The power of the heater is given as \(500 W\). The heat energy (\(Q\)) supplied by the heater during this time can be calculated using the formula:
\[ Q = \text{Power} \times \text{Time} = 500 \, W \times 16 \, s = 8000 \, J \]

### Step 3: Relate heat energy to heat capacity
The heat energy supplied to the block is also related to the heat capacity (\(H_c\)) and the temperature change:
\[ Q = H_c \times \Delta T \]
Substituting the values we have:
\[ 8000 \, J = H_c \times 40^\circ C \]

### Step 4: Solve for heat capacity
Now, we can solve for the heat capacity (\(H_c\)):
\[ H_c = \frac{8000 \, J}{40^\circ C} = 200 \, J/^\circ C \]

### Final Answer
The heat capacity of the block is \(200 \, J/^\circ C\).

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