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Ilia_Sergeevich [38]
3 years ago
5

A balloon has a maximum capacity of 1700 m3. On a cool morning (temperature = 10°c, some hot-air balloonists start filling their

balloon with air until it occupies three-fourths of the maximum possible volume. Then, they turn on the propane burner to fill the balloon to its complete volume of 1700 m3. The temperature at which the air fills the balloon to its maximum capacity is:
Chemistry
1 answer:
bulgar [2K]3 years ago
7 0

V1/T1 = V2 / T2

V1 = 3/4(1700) = 1275

T1 = 10 + 273.15 = 283.15

V2 = 1700

T2 = V2T1

T2 = (1700) (283.15) / 1275

T2 = 377.53 K

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Answer:

T_2=17.8\°C

Explanation:

Hello,

In this case, we can solve this problem by noticing that the heat lost by the warm water is gained by the ice in order to melt it:

Q_{water}=-Q_{ice}

In such a way, the cooling of water corresponds to specific heat and the melting of ice to sensible heat and specific heat also that could be represented as follows:

m_{water}Cp_{water}(T_2-T_{water})=-m_{ice}\Delta H_{melting,ice}-m_{ice}\Cp_{ice}(T_2-T_{ice})

Thus, specific heat of water is 4.18 J/g°C, heat of melting is 334 J/g and specific heat of ice is 2.04 J/g°C, thus, we can compute the final temperature as shown below:

m_{water}Cp_{water}(T_2-T_{water})+m_{ice}Cp_{ice}(T_2-T_{ice})=-m_{ice}\Delta H_{melting,ice}\\\\T_2=\frac{-m_{ice}\Delta H_{melting,ice}+m_{water}Cp_{water}T_{water}+m_{ice}Cp_{ice}T_{ice}}{m_{water}Cp_{water}+m_{ice}Cp_{ice}} \\\\T_2=\frac{-50.0*334+100*4.18*67+50.0*2.04*-20.0}{100*4.18+50.0*2.04} \\\\T_2=17.8\°C

Best regards.

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