A 4.10 kg silver ingot is taken from a furnace, where its temperature is 747degrees C, and placed on a very large block of ice at 0.00 degrees C. Assuming that all the heat given up by the silver is used to melt the ice and that not all the ice melts, how much ice is melted?
2007-12-06 18:02:56 · 4 answers · asked by ! 2 in Science & Mathematics ➔ Physics
Use the formula Q = m_s*c*ΔT to find the amount of heat needed to be transfered from the silver ingot to lower it's temperature to zero degrees celcius. The specific heat (c) for silver is 234 J/kg*K.
Use the formula Q = m_w*L_f to find the amount of ice melted. The heat of fusion (L_f) for water is 4190 J/kg*K.
Since the amount of heat transfered is from the silver to the water, Q is the same for both equations.
so, m_s*c*ΔT = m_w*L_f
solve for m_w
m_w = m_s*c*ΔT /L_f
m_w = 4.10 * 234 * 747 / 4190 = 171 kg
2007-12-06 19:38:50 · answer #1 · answered by Demiurge42 7 · 0⤊ 2⤋
The specific heat of silver is needed to solve this. If that number is S calories per gram per degree, then the number of calories delivered to the ice is 747 x 4100 S, and the number of calories absorbed by melting ice is 747 x 4100 S/80.
2007-12-06 18:10:07 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The specific heat of silver is needed to solve this. If that number is S calories per gram per degree, then the number of calories delivered to the ice is 747 x 4100 S, and the number of grams of ice melted is 747 x 4100 S/80.
Postscript: the answer following this one is incorrect. The responder copied my original text, which had an error which I have fixed. I am going to report the answer for plagiarism.
2007-12-06 18:08:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋
that doesnt make sense? whats the exact size of the ice block?
2007-12-06 18:05:40 · answer #4 · answered by Anonymous · 0⤊ 1⤋