A 195-g sample of a metal was heated to 120.0 degrees C, and then placed into 200.0g water at 21.8 degrees C. The final temperature of the water was measured to be 52.5 degrees C. Calculate (a) the heat transferred from the metal to water and (b) the specific heat of the metal. (Assume no heat loss to the surroundings)can someone please show me HOW to do this
specific heat of water 4.184 J
specific heat of copper 0.385 J is also given in the problem
thankyou
specific heat of water 4.184 J
specific heat of copper 0.385 J is also given in the problem
thankyou
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Heat Transferred to the water (This was a gain).
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m = 200.0 grams
c = 4.184 J/g*oC
delta t = 52.5 - 21.8 = 30.7oC
Heat = mc delta(t)
Heat = 200.0 * 4.184 * 30.7 = 25700 J to 3 places. Since this is an answer, I'll right to the correct number of significant digits, but we should use 25689.8 in the calculation below.
Heat Lost
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m = 195 grams
c = ????
delta t = 120.0 - 52.5 = 67.5
Heat Lost = 195*c * 67.5 = 13163 which I won't fully round quite yet.
Heat Lost = heat gained.
13163 c_metal = 25689.8
c_metal = 1.9516 which to three places is
c_metal = 1.95
I don't know what the copper has to do with it. This value seems high for a metal.
=====================
m = 200.0 grams
c = 4.184 J/g*oC
delta t = 52.5 - 21.8 = 30.7oC
Heat = mc delta(t)
Heat = 200.0 * 4.184 * 30.7 = 25700 J to 3 places. Since this is an answer, I'll right to the correct number of significant digits, but we should use 25689.8 in the calculation below.
Heat Lost
=======
m = 195 grams
c = ????
delta t = 120.0 - 52.5 = 67.5
Heat Lost = 195*c * 67.5 = 13163 which I won't fully round quite yet.
Heat Lost = heat gained.
13163 c_metal = 25689.8
c_metal = 1.9516 which to three places is
c_metal = 1.95
I don't know what the copper has to do with it. This value seems high for a metal.