ice_example - Q 1 = c ice m (40 C ), where c ice = 2220 J /...

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A 1 kg block of ice is initially at - 40 F. It is placed in a container which holds 10 kg of water at 80 C. Assuming the container is perfectly insulating (i.e., the water+ice are completed isolated), what will be the ±nal temperature of the water after the ice melts? ========================================= Since this system is completely isolated, no heat can enter or leave. Thus, the heat released by the 10 kg of water as it cools from 80 C to the ±nal temperature T F must equal the total heat involved in: (1) warming the ice to the melting temperature (0 C), (2) melting it, and (3) warming the 1 kg of water from 0 C to the ±nal temperature T F . The heat released is Q = c water M ( T F - 80 C), where c water = 4190 J / kg · C is the speci±c heat of water and M = 10 kg is the mass of the warm water. The heat required to warm the ice is
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Unformatted text preview: Q 1 = c ice m (40 C ), where c ice = 2220 J / kg C is the specic heat of ice and 1 = 1 kg is the mass of the ice. The temperature change is 40 C , since-40 F =-40 C. So, Q 1 = 88 . 8 kJ. The heat required to melt the ice is Q 2 = L F m , where L F = 333 kJ / kg is the heat of fusion for ice. So, Q 2 = 333 kJ. The heat required to warm this water to the nal temperature is Q 3 = c water m ( T F- C). Since Q + Q 1 + Q 2 + Q 3 = 0, c water M ( T F-80 C) + Q 1 + Q 2 + c water mT F = 0 . One simply solves for T F . T F = c water M (80 C)-Q 1-Q 2 c water ( M + m ) T F = (4190 J / kg C )(10 kg)(80 C)-88 . 8 kJ-333 kJ (4190 J / kg C )(10 kg + 1 kg) = 64 C...
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