If a 50-gram piece of metal that has a temperature of 110 degrees celsius was dropped into 1000 grams of water with the temperature of 25 degrees celcius, what would happen?

a) the water will quickly reach its boiling point
b) the water's temperature will stay the same but the metal will cool down
c) the water's temperature will increase and the metal will stay constant
d) the water and the metals temperature will reach the same temperature

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Ответ:

karenjjensen

12.06.2022

Chemistry

The tabulated values at 120 C of hfg = 2202.1 kJ/Kg and sfg = 5.6

kJ/Kg.K

Explanation:

According to Clapeyron Equation :

$h_{fg}=TV_{fg}\left (\frac{\partial P}{\partial T} \right )_{sat}$

$=T(V_{g}-V_{f})_{120^{0}}\left ( \frac{\Delta P}{\125^{0}-115^{0}} \right )$

Taking P2 at temperature 125 and P1 at 115 degree Celsius .

T = 120 + 273.15 = 393.15 K

P2 = 232.23 kPa and P1 = 169.18kPa

(these are experimental values given in the Data)

$= 393.15(0.88133 -0.001060\ m^{3}/Kg)\frac{232.23-169.18 kPa}{10}$

$h_{fg} = 2206.8 kJ/Kg$

$S_{fg}=\frac{h_{fg}}{T}$

$\frac{2206.8}{393.15}$

= 5.6131 kJ/Kg.K

The tabulated values at 120 C of hfg = 2202.1 kJ/Kg and sfg = 5.6131

kJ/Kg.K

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