An unknown substance with a mass of 13.00 g is heated up to 75.0 °C and then add

Question

An unknown substance with a mass of 13.00 g is heated up to 75.0 °C and then added to a coffee cup calorimeter containing 25.0 g of water at 21.3 °C. If the temperature of the water increases to 27.7 °C, what is the specific heat of the unknown? The specific heat of liquid water is 4.184 J/g°C

How many joules of energy is required to convert 125 g of water at 21 °C into steam at 110 °C? The specific heat of liquid water is 4.184 J/g°C; the specific heat of steam is 2.09 J/g°C; the latent heat of vaporization of water is 2260 J/g. Assume the boiling point of water is 100 °C.

Explanation / Answer

part1)

heat given by substance = heat absorbed by water

heat given by substance = - mass x specific heat x rdrop in temperature

= - 13.0g x s x (27.7-75)

= 614.9s J

heat absorbed by water = 25.0g x 4.184J/g.k x (27.7-21.3)

= 669.44 J

Equating both s = 669.44/614.9

= 1.08869 J /g.K

Part 2)

H2O (l) -----------> H2O(l) ------------> H2O(g) -------------> H2O(g)

21C q1 0C q2 0C q3 110C

q1 = mass x specific heat of water x rise in temperature

= 125g x 4.184J/g.K x 79 K

= 41317 J

q2 = mass x latent heat of vaporization

= 125 g x 2260J/g

= 282500 J

q3 = mass x specific heat of steam x rise in temperature

= 125 x2.09J/g.K x 10K

= 2612.5 J

Thus the total heat required to heat 125 og water to steam at 110C is

Q = q1 +q2 + q3

= 41317 J +282500 J +2612.5 J

=326429.5 J

=326.43kJ

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