300 grams of led at 200 degrees celcius is dropped into 150 grams of water at 10 degrees celcius. if the final temperature is 20 degrees celcius, calculate the specific heat capacity of the led. (specific heat capacity of water is 4200 joules per kilogram degrees celcius
sum of heats gained is zero (some lose heat).
heat gained by lead + heat gained by water =0
300*cPb*(20-200)+ 150*cWater*(20-10)=0
solve for cPb (cwater= 4.2j/gramC)