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7AW11Quiz2ASolution - fnal mass oF liquid water= initial...

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Lecture A 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 -27° C ) Δ Eth (ice) = m(ice) c(ice) Δ T= 0.8kg 2.06kJ/(kg K) (0- -27) = 44.5 kJ Water( ±rom 10 C) Δ Eth ( liq) = m(liq) c(liq) Δ T= 0.5 kg 4.18 kJ/( kg K) ( 0-10) = -20.9kJ 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 44.5 kJ - 20.9kJ - Δ m (333.5 kJ/kg) = 0 Δ m = 0.071kg o± liquid water will change into solid. There±ore at equilibrium : fnal mass oF ice = initial mass o± ice + Δ m = 0.8kg + 0.071kg= 0.871kg
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Unformatted text preview: fnal mass oF liquid water= initial mass o± liq- Δ m= 0.5kg - 0.071kg= 0.429kg Problem 2) Car A will have a larger compression on the spring, since both cars have the same speed but di±±erent masses and mass A> mass B => Δ KE ( car A) > Δ KE ( car B), where as we know Δ KE = 1/2 m Δ v ² . And considering the system is closed, whichever car has greater change in KE will have the greater change in PE as well, so Δ PE (car A) > Δ PE (car B). With Δ PEs = 1/2 k Δ x ² , which means that Δ x ( car A) > Δ x (car B). Ice Water Water Eb m (liq) Eth T Eth T Ti= -27°C Ti= 10° C mi= 0.5kg T±= 0°C T±= 0° C m±= 0.429kg...
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