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3 years ago

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179.1 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 16.1oC. After 306.9 g

of an unknown compound at 94.3oC is added, the equilibrium T is 27.4oC.
What is the specific heat of the unknown compound in J/(goC)?

1 answer:

taurus [48]3 years ago

4 0

Answer:

the specific heat of the unknown compound is $c_u=0.412J/g \cdot C$

Explanation:

Generally the change in temperature of water is evaluated as

$\Delta T = T_2 -T_1$

Substituting 16.1°C for $T_1$ and 27.4°C for $T_2$

$\Delta T = 27.4 - 16.1$

$=11.3^oC$

Generally the change in temperature of unknown compound is evaluated as

$\Delta T_u = T_3 -T_2$

Substituting 27.4°C for $T_2$ and 94.3°C for $T_3$

$\Delta T = 94.3 - 27.4$

$=66.9^oC$

Since there is an increase in temperature then heat is gained by water and this can be evaluated as

$H_w = mc_w \Delta T$

Substituting 179.1 g for m , 4.18 J/g.C for $c_w$ (specific heat of water)

$H_w = 4.18 * 179.1 * 11.3$

$= 8459.6J$

Since there is a decrease in temperature then heat is lost by unknown compound and this can be evaluated as

$H_u = m_uc_u \Delta T_u$

By conservation of energy law

Heat lost = Heat gained

Substituting 306.9 g for $m_u$ , 8459.6J for $H_u$

$8459.6 = 306.9 * c_u * 66.9$

Therefore $c_u = \frac{8459.6}{308.9 *66.9}$

$=0.412J/g \cdot C$

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