Specific Heat Problem?

Heat lost by the metal = Heat gained bythe water.

mcΔT = Q: Where m = mass; c = specific heat: ΔT = temp. change: Q = Heat absorbed or released.

Heat lost by the Lead:

= 200g x c (cal/g/°C) x 98.81 °C ΔT = 19,762 x c cal/g/°C

Heat gained by the water.

= 500.0g x 1.0 cal/g/°C x (11.19°C -10 ) = 500 g x 1.19°C ΔT = 595 cal

Heat lost = heat gained

c x 19,762 = 595 cal

c (SH of lead = 595 / 19762 = 0.03 cal/g/°C

(0.03 x 4.184 J/g/°C = 0.126 J/g/°C).

Keith - it is been a real long term for me because chemistry... and that i'm, needless to say, "winging" this: to strengthen the temperature of the 250 g of water by ability of 1ºC demands 250 x 4.184 = a million,046 J. To get that plenty warmth out of 155g of Cu the copper will would desire to cool (a million,046)/(.385 x one hundred fifty five) = 17.53ºC the merely precise temperature... of path... would be T. The copper cools 17.fifty 3 cases as plenty via fact the water warms the version between the nice and cozy copper and the cool water is, initially, one hundred thirty.2 C stages. i'm questioning of the temperature hollow ultimate in increments of 18.fifty 3 C stages. it relatively is by the fact, for each 1º strengthen in water temp., there's a 17.53º decrease interior 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 strengthen; water temp. ends at 26.8º. permit's examine: If the copper cooled an entire of seven x 17.53º it is going to cool approximately 123.2º. that would have the copper end at one hundred fifty - 123.2 = 26.8º sure! do no longer you in user-friendly terms like it while it works? Richard

heat loss by metal=heat gained by water.

C=Q/temperature change

C=heat capacity

Q=amount of thermal energy transferred in Joules

C=mc(change in temperature)

m=mass

c=specific heat capacity

500 x 1.19 = 595 cal

110 = 11.9 = 98.1 deg C

595 cal / (200 ) = 2.975 cal / gram

2.975 / 98.1 = .03 cal / g / deg C

This doesn't look right.