A piece of unknown metal with a mass of 23.8g and is heated to 100 degrees C. and it is dropped into 50 mL of water at 24 degrees C. Final temp of system is 32.5 degrees C. What is metal's specific heat in joules/g degree celcius?
2006-10-11 14:25:53 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
the key is that heat lost by the metal is gained by the water
heat(gained by water)=mass of water*change in temperature of water*specific heat of water. The specific heat of water is 4.18J/(gC). 50ml water = 50g water
so plug in the values and find the heat that the water gained.
This value for heat = the amount of heat that the metal lost
so... use the same equation as above except now for the metal
heat(lost by metal)=mass of metal*change in temperature of the metal*specific heat of metal
plug in the numbers and solve for the specific heat
2006-10-11 14:32:55 · answer #1 · answered by Greg G 5 · 0⤊ 0⤋
2
2006-10-11 14:28:04 · answer #2 · answered by Mark Antony 3 · 0⤊ 0⤋
The heat content of the metal is 23.8*100*cpm (cpm is the specific heat of the metal). Heat content of the water is 50*24*cpw.
The total heat quantity is 23.8*100*cpm + 50*24*cpw before the parts are combined. After they are combined; the total heat quantity is 23.8*32.5*cpm + 50*32.5*cpw. These must be equal, therefore:
23.8*100*cpm + 50*24*cpw = 23.8*32.5*cpm + 50*32.5*cpw
cpw is 4.15 J/(gm ºC). Solve for cpm
2006-10-11 14:40:01 · answer #3 · answered by gp4rts 7 · 0⤊ 0⤋
Keith - it somewhat is been a actual lengthy time for me considering chemistry... and that i'm, of course, "winging" this: to strengthen the temperature of the 250 g of water by utilising 1ºC demands 250 x 4.184 = one million,046 J. To get that lots warmth out of 155g of Cu the copper will would desire to cool (one million,046)/(.385 x a hundred and fifty five) = 17.53ºC the terrific temperature... of direction... would be T. The copper cools 17.fifty 3 instances as lots because of the fact the water warms the adaptation between the warm copper and the cool water is, initially, one hundred thirty.2 C levels. i'm questioning of the temperature hollow ultimate in increments of 18.fifty 3 C levels. that's using the fact, for each 1º enhance in water temp., there's a 17.53º shrink in the copper's temp... 18.53º closure. one hundred thirty.2/18.fifty 3 = 7.02. there'll be 7 of those 1º water temp. will enhance; water temp. ends at 26.8º. enable's verify: If the copper cooled a comprehensive of 7 x 17.53º it is going to cool approximately 123.2º. that would desire to have the copper end at a hundred and fifty - 123.2 = 26.8º sure! do not you purely like it whilst it works? Richard
2016-12-16 06:16:49 · answer #4 · answered by coman 3 · 0⤊ 0⤋
Your mama! hehe
2006-10-11 14:33:16 · answer #5 · answered by Anonymous · 0⤊ 0⤋