a 102 kilogram block of aluminum at T=60 degree C is submerged in 900.0g of water at T= 5 degree C. Following heat transfer, what is the final temperature of the water and aluminum (s=0.900j/g degree C for aluminum

Thank you for the opportunity to help you with your question!

specific heat of water =4.186

specific heat of aluminium=0.900

from the given condition,

102*1000*0.9*(60-T)=900*4.186*(T-5)

solving for T we get,

5.5 *10^6-91800T=3767.4T-18837

95567.4T=5.52*10^6

T=57.83 C

1.2*1000*0.9*(60-T)=900*4.186*(T-5)

64800-1080T=3767.4T-18837

4847.4T=83637

T=17.25 C

where did you get the 4.186 from

A block of copper at T=12 degree c loses 30000 J of heat and it's temperature drops to -10 degree c what is the mass of the block s= 0.385 j/g c

Q=cmdt

30000=0.385*m*22

m=30000/8.47

m=3541 g

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