Question

Any formulas Shown along with work would be greatly appreciated!

7. A 0.500 kg glass (C=840 J/kg°C) containing 1.00 L of water (at 20.0°C) is filled with 0.100 kg of ice (at -5.0°C). The specific heat of ice is 2108J/kg K a. What is the mass of the liquid water initially in the glass? b. What is the equilibrium temperature of the water and glass when all of the ice has melted? Ignore any heat gained from or lost to the surroundings. C. A person then drinks all of the water in the glass. How much heat is gained by the water as its temperature increases to 37.0°C in the digestive tract of the person?

Answer 1

a)

Volume of liquid water $V=1\,L=10^{-3}\,m^3$

Density of water $d=1000\,kg/m^3$

Mass of liquid water in glass $m_w=dV=1000*10^{-3}=1\,kg$

b)

Heat lost by water+glass = Heat gained by ice

$m_gC_g(20-T)+m_wC_w(20-T)=m_iC_i(0-(-5))+m_iL_i$

$(0.500*840+1*4186)(20-T)=0.100*2108(5)+0.100*334*10^3$

$92120-4606T=34454$

$T=12.52\degree$

Equilibrium temperature is $T=12.52\degree$

c)

Heat gained by digestive tract of person is $H=(m_w+m_i)C_w(37-12.52)$

$H=(1.100)4186(37-12.52)=112722\,J$