if 655g of diamond is immersed in 2.5 L of water (assume the density of water...

if 655g of diamond is immersed in 2.5 L of water (assume the density of water is 1g/mL) initially at room temperature (25 degree celsius) and the sample is initially at 815 degree celsius, determine the final temperature of both water and diamond

Expert Solution

specific heat capacity of diamon = 0.51J/g-⁰C

mass of water = volume * density

= 2500*1 = 2500g

Heat lose of diamond = Heat gain of water

mcT = mcT

655*0.51*(815-t) = 2500*4.184*(t-25)

t = 49.45⁰C

The final temperature of both water and diamond = 49.45⁰C

queen_honey_blossom answered 2 years ago

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