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Question of the Day
An unknown metal of 50g is initially at 100°C and dropped into a beaker of 200g of water at 30°C. The final temperature of the system is 40°C. Find the specific heat of the metal of the specific heat of water is 4.2 J/g°C.
Heat lost by metal= heat gained by water mcΔT we have 50x c x (100-40) = 200 x 4.2 x (40-30) so c = 2.8J/g°C
2.8J/g°C
4.2J/g°C
5.1J/g°C
6.4J/g°C
10J/g°C
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