A calorimeter has mass 100g and specific heat `0.1 kcal//kg^(@)C` . It contains 250 g of liquid at `30^(@)C` having specific heat of `0.4 kcal//kg^(@)C` . If we drop a piece of ice of mass 10 g at `0^(@)C` , what will be the temperature of the mixture ?

Answer

Temperature of the mixture will be `20.8^@C`

Explanation

Let T be the final temperature attained by water (turned from ice), calorimeter and liquid. We have

We know that: Latent heat of melting the ice `= L_("melt")`
Calculation:
Heat gained in converting ice to water at `0^@C, Q_3 =m_("ice")L_("melt")=10xx80 =800 ` cal .
Heat gained in raising temperature of water from `0^@ " to " T^@C,`
`Q_2 = m_("water")C_("water") Delta T`
`=10 xx 1 xx (T-0)= 10 T`
Heat lost by liquid cooling from `30^@C "to" T^@C`,
`Q_3=m_("liq") C_("liq") Delta T' `
`=250 xx 0.4xx (30 -T) `
Heat lost by calorimeter in cooling from `306@C " to " T^@C`
`Q_4 =m_("calorimeter ")C_("calorimeter")Delta T,`
`=100 xx 0.1 xx (30-T)=10 xx (30-T)`
According to principle of heat exchange,
`Q_1 +Q_2=Q_3+Q_4`
`therefore 800 + 10T =100(30-T)=10(30-T)`
`therefore 800 +10 T =110 (30-T)`
`therefore 800 +10 T=3300 -1100 T`
`therefore 120 T = 300-800=2500`
`therefore T= (2500)/(120) = 20.8 ^@C`