A metal block absorbs `4500 cal` of heat when heated from `30^@C` to `80^@C`. Its thermal capacity is

Options

A. 90 gm
B. `90 cal//^(0) C`
C. 9 gm
D. `9 cal//^(0) C`

Explanation

To find the thermal capacity of the metal block, we can follow these steps:

### Step 1: Identify the given values
- Heat absorbed (Q) = 4500 cal
- Initial temperature (T1) = 30°C
- Final temperature (T2) = 80°C

### Step 2: Calculate the change in temperature (ΔT)
The change in temperature (ΔT) can be calculated using the formula:
\[ \Delta T = T2 - T1 \]
Substituting the values:
\[ \Delta T = 80°C - 30°C = 50°C \]

### Step 3: Use the formula for thermal capacity (C)
The formula for thermal capacity (C) is given by:
\[ C = \frac{Q}{\Delta T} \]
Where:
- C = thermal capacity
- Q = heat absorbed
- ΔT = change in temperature

### Step 4: Substitute the values into the formula
Now we can substitute the values of Q and ΔT into the formula:
\[ C = \frac{4500 \text{ cal}}{50 \text{ °C}} \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ C = \frac{4500}{50} = 90 \text{ cal/°C} \]

### Final Answer
The thermal capacity of the metal block is **90 cal/°C**.

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