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A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celsius is added to...

A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celsius is added to 50.00 grams of water initially at 12.16 degrees Celsius. The final temperature of the metal and water is 20.15 degrees Celsius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C).

(HINTS: Recall that the q of a system is equal to the -q of the surroundings. So you can set up two equations here: one for the loss of heat from the metal (the system), and one for the gain of heat by the water (the surroundings). Then substitute into one equation the variables from the other equation as appropriate. Additionally, the final temperature of the metal and the water must be the same temperature because the metal is in the water when they reach a final temperature.)

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Answer #1

Answer -

Given,

Mass of unknown sample = 25.00 g

Initial Temperature of unknown metal = 99.0 \degree C

Mass of water = 50.00 g

Initial Temperature of water = 12.16\degreeC

Final temperature of both = 20.15 \degree C

Specific heat of water = 4.184 J/g.\degreeC

Specific heat of unkown metal = ?

We known that,

Heat = m * s * \Delta T

where, m = mass

s = specific heat

\DeltaT = change in temperature (Final - initial)

Change in Temperature of water = 20.15 \degree C - 12.16\degreeC = 7.99 \degree C

Change in Temperature of unkown metal = 20.15 \degree C - 99.0 \degree C = -78.85\degreeC

Now,

there is not heat loss, So, Heat released by metal is absorbed by water.

- Heatm = Heatw

mm * sm * \Delta Tm = mw * sw * \Delta Tw

Put the values,

- (25.00 g * sm * -78.85\degreeC) = 50.00 g * 4.184 J/g.C * 7.99 \degree C

sm = (50.00 g * 4.184 J/g.\degreeC * 7.99 \degree C) /[-(25.00 g * -78.85\degreeC)]

sm = 0.848 J/g.\degreeC [Answer]

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