In an experiment on the specific heat of a metal a 0.20 kg block of the metal at `150^(@)`C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cc of water at `27^(@)`C. The final temperature is `40^(@)`C. Calcualte the specific heat of the metal. If heat losses to the surroundings are not negligible, is our answer greater or smaller than the actual value of specific heat of the metal?

Explanation

Given that the block of metal is dropped into a calorimeter,
Mass of metallic block,
m = 0.20 kg = 200 g
As the block is dropped into calorimeter, its temperature falls to `40^@C` , that means it loses heat energy and same amount of heat energy is gained by calorimeter.
Change in temperature of metal block,
` Delta T_b = 150 - 140 = 110^@C`
Change in temperature of calorimeter,
` Delta T_C = 40 - 27 = 13^@C`
Mass of water in calorimeter = 0.025 kg
= 25 g
Water equivalent of the calorimeter = 150 g [ `because ` volume of water = 150 cc]
So, we can consider that the calorimeter behaves as a (150 + 25) of water.
Let the specific heat of metal block be C. Specific heat of water,
`C. = 4.2 J//g ""^@C`
Heat energy lost by metal block = Heat energy gained by calorimeter.
`2m_b = C Delta T_b= m_w C. Delta T_C`
` rArr C = (175 xx 4.2 xx 13)/(200 xx 110) = 0.434 J//g ""^@C`