Question

A liquid of unknown specific heat at a temperature of 20oC was mixed with water at 80oC in a well-insulated container. The final temperature was measured to be 50oC, and the combined mass of the two liquids was measured to be 240g. In a second experiment with both liquids at the same initial temperatures, 20 g less of the liquid of unknown specific heat was poured into the same amount of water as before. This time the equilibrium temperature was found to be 52oC. Determine the specific heat of the liquid

Answer 1

Answer:

Specific heat capacity of unknown liquid is given as

$s_{liq} = 7.77 J/g ^oC$

Explanation:

Let the mass of unknown liquid is "m"

Now when unknown liquid is mixed with water then heat given by water is equal to the heat absorbed by unknown liquid

So we have

$m s_{liq} (50 - 20) = (240 - m)(4.2)(80 - 50)$

Now again we did the same experiment but this time the mass of unknown liquid is "m - 20"

so we have

$(m - 20) s_{liq} (52 - 20) = (220 - m)(4.2)(80 - 52)$

now from above two equations

$\frac{m(30)}{(m - 20)32} = \frac{(240 - m)30}{(220 - m)28}$

Now we have

$30 m \times 28 (220 - m) = 32(m - 20) \times 30 (240 - m)$

so we have

$m = 84.2 g$

Now we have

$s_{liq} = \frac{240 - 84.2}{84.2} (4.2)$

$s_{liq} = 7.77 J/g ^oC$

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Topic : Calorimetry

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