Question

In an experiment to determine 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 cm^3` of water at `27 .^(@) C`. The final temperature is `40.^(@) C`. The specific heat of the metal is.

Answer 1

Given mass of metal `M=0.20kg=200g`
Total mass of calorimeter`=(m+w)=(150c cxx1g//c c+0.025xx1000)g`
(Density of water`=1g/c c`)
Fall in temperature of metal=150-40=`110^(@)C`
Rise in temperature of water=40-27=`13^(@)C` ltBrgt heat lost be metal=MC(110)
=22000C
Heat gained by calorimeter `=(m+w)xx1xx(13)`
`=2275cal` (specific heat capacity of water=1 cal `g^(-1).^(@)C^(-1))`
As heat lost=heat gained
`therefore22000C=2275`
Solving we get C=0.1cal/gm`.^(@)C`