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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 speAny formulas Shown along with work would be greatly appreciated!

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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

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