A calorimeter has mass 100 g and specific heat 0.1 kcal/kg ""^@C. It contains 250 g of liquid at 30 ""^@Chaving 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?

A calorimeter has mass 100 g and specific heat 0.1 kcal/kg ""^@C. It c

1.	A calorimeter has mass 100 g and specific heat 0.1 kcal/kg ""^@C. It contains 250 g of liquid at 30 ""^@Chaving 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» Solution :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` `therefore120 T = 300-800=2500` `thereforeT= (2500)/(120) = 20.8 ^@C`