what is the final temperature of 30.0 g pure sample of liquid propanone after ad

Question

what is the final temperature of 30.0 g pure sample of liquid propanone after adding 30.6 kJ of energy to the sample initally at -110 c?

the answer i think should be 78.5 please explain and put the steps

thank you

c. melting a sample of liquid water at its normal melting point of UL. U d. vaporizing a sample of liquid water to gaseous water at its normal boiling point. 2. What is the final temperature of a 30.0g pure sample of liquid propanone after adding 30.6 kJ of energy to the sample initially at –110.0°C? Specific heat (solid), 1.66 J/gk | Melting point, -94.7°C Heat of vaporization, 31.3 kJ/mole Specific heat (liquid), 2.16 J/gK Boiling point, 56.1°C Heat of fusion, 5.7 kJ/mole Specific heat (gas), 1.3 J/gK Molar mass, 58.0 g/mol 35.0 °C (b. 78.5 °C 115.5 °C 370.0 °C Mošo Veso-| T - lo Fe 600 - (30)( 2 KG) (T* H (... (Tel63

Explanation / Answer

Q = mcT

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kgK), is a symbol meaning "the change in"

T = change in temperature (Kelvins, K)

The Liquid Praponal is undergoing following changes

1) It reahces -94.7 Deg Cel From -110 Deg Cel

2) Then It melts at -94.7 Deg Cel

3) It reahces 56.1 Deg Cel From -94.7 Deg Cel

4) Then it Converts to vapour

5) Accroding to the energy it reaches some temperature

Then we do calculation here

[ (110-94.7) x 30 x 1.66 ]+ [(30/58) x 5700 ]+ [(94.7+56.1) x 30 x 2.16] +  [(30/58) x 31300 ] + [(X-56.1) x 30 x 1.3 ] = 30600

761.94 + 2948.275 + 9771.84 + 16189 + 39X - 2187.9 = 30600

If you calculate X

X = 78.5 Deg Cel

Hence Option C is correct answer

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