Question

A piece of lead of mass 112 grams is heated to 100 degree celcius and dropped into a copper calorimeter of mass 40 grams containing 200 grams of water at 16 degree celsius. Neglecting the heat loss, find the specific heat of the lead if the resultant temperature attained is 24.1 degree celcius(specific heat of copper is 0.09 cal/g degree celcius) ​

Answer 1

Specific heat of the lead if the resultant temperature attained is 24.1 °C is 0.194 cal/g/°C

Given-

• Mass of lead piece = 112 grams
• Temperature of lead = 100 °C
• Mass of copper calorimeter = 40 grams
• Weight of water = 200 grams
• Temperature of water = 16 °C
• Resultant temperature = 24.1 °C
• Specific heat of copper = 0.09 cal /g /°C

We know that principle of calorimeter is heat released is equals to the heat absorbed.

Heat will be released by the metal. So

Heat released = 112 × S × (100 - 24.1) = 112 × S × 75.9

S is the specific heat which we have to find

Heat absorbed = 40 × 0.09 × (24.1 - 16) + 200 × 0.09 × (24.1 - 16)

Heat absorbed = 40 × 0.09 × 8.1 + 200 × 8.1 = 1649.16

Heat released = heat absorbed

112 × S × 75.9 = 1649.16

S = 0.194 cal/g/°C