7AW11Quiz2BSolution

7AW11Quiz2BSolution - initial mass o liq- m= 0.4kg -...

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Lecture B Problem 1) First we calculate to see which one will go a phase change frst. So we see which one will reach 0° C. Ice ( ±rom -22° C ) Δ Eth (ice) = m(ice) c(ice) Δ T= 0.6kg 2.06kJ/(kg K) (0- -22) = 27.1 kJ Water( ±rom 10 C) Δ Eth ( liq) = m(liq) c(liq) Δ T= 0.4 kg 4.18 kJ/( kg K) ( 0-10) = -16.7kJ Which means that liquid water will undergo the phase change frst. Δ Eth(ice) + Δ Eth(water) + Eb(water) = 0 m(ice) c(ice) Δ T(ice) + m(liq) c(liq) Δ T(liq) + Δ m Δ H = 0 27.1 kJ - 16.7kJ - Δ m (333.5 kJ/kg) = 0 Δ m = 0.031kg o± liquid water will change into solid. There±ore at equilibrium : fnal mass oF ice = initial mass o± ice + Δ m = 0.6kg + 0.031kg= 0.631kg fnal mass oF liquid water=
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Unformatted text preview: initial mass o liq- m= 0.4kg - 0.031kg= 0.369kg Problem 2) Rock B will reach a higher maximum height ater releasing, since they are launched rom identical springs rom the same displacement- so PEs (rock B) = PE s (rock A), which means they are launched with same amount o Kinetic energy, KE are equal. Also both rocks will have same PEg while reaching their max heights, considering the systems closed. PEg(rock B) = PEg(rock A), but with PEg= mg h and mA > mB => h(rock B) > h(rock A). Ice Water Water Eb m (liq) Eth T Eth T Ti= -22C Ti= 10 C mi= 0.4kg T= 0C T= 0 C m= 0.369kg...
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