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.)
Answer -
Given,
Mass of unknown sample = 25.00 g
Initial Temperature of unknown metal = 99.0 ^{ }C
Mass of water = 50.00 g
Initial Temperature of water = 12.16^{}C
Final temperature of both = 20.15 ^{ }C
Specific heat of water = 4.184 J/g.^{}C
Specific heat of unkown metal = ?
We known that,
Heat = m * s * T
where, m = mass
s = specific heat
T = change in temperature (Final - initial)
Change in Temperature of water = 20.15 ^{ }C - 12.16^{}C = 7.99 ^{ }C
Change in Temperature of unkown metal = 20.15 ^{ }C - 99.0 ^{ }C = -78.85^{}C
Now,
there is not heat loss, So, Heat released by metal is absorbed by water.
- Heat_{m} = Heat_{w}
m_{m} * s_{m} * T_{m} = m_{w} * s_{w} * T_{w}
Put the values,
- (25.00 g * s_{m} * -78.85^{}C) = 50.00 g * 4.184 J/g.C * 7.99 ^{ }C
s_{m} = (50.00 g * 4.184 J/g.^{}C * 7.99 ^{ }C) /[-(25.00 g * -78.85^{}C)]
s_{m} = 0.848 J/g.^{}C [Answer]
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