A calorimeter of water equivalent 100gm cools in air in 18 minutes from `60^(@)C` to` 40^(@)C`.When a block of metal of mass 60 gm is heated to `60^(@)C` and placed inside the calorimeter. Assume heat loss only by radiation and Newton's Law of cooling to be valid. Find the specific heat of metal if now the system cools from `60^(@)C` to `40^(@)C` in 20 minutes.

Explanation

To solve the problem step by step, we will use Newton's Law of Cooling and the concept of heat transfer.

### Step 1: Understanding the Initial Cooling of the Calorimeter
The calorimeter cools from 60°C to 40°C in 18 minutes. We can denote:
- \( T_2 = 60°C \) (initial temperature)
- \( T_1 = 40°C \) (final temperature)
- \( T_s \) = surrounding temperature (unknown)
- \( t = 18 \) minutes
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