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-A spherical balloon is inflated to a diameter of 20.0 cm. Assume that the gas in the balloon is of atmospheric pressure (101.3 kPa) and is at a temperature of 20.0 °C. It is then taken by a diver 15.0 m under the sea. The temperature of the seawater at this depth is 16.0 °C . Density of seawater: 1030. kg m–3. The gauge pressure of the air in the balloon at this depth is 27300Pa.
Q1>Assuming the gas in the balloon is in thermal equilibrium with seawater, what is the volume of the balloon now?

-Steam at 100 °C is mixed with 166.4 g of ice at –32.8 °C, in a thermally insulated container, to produce water at 44.6 °C. Ignore any heat absorption by the container.
-Cwater = 4186. J/(kg °C)
-Cice = 2090. J/(kg °C)
-Lf,water = 3.33 × 105 J/kg
-Lv,water = 2.26 × 106 J/kg
Also,energy is required to bring all the ice up to 0 °C=11400J;
energy is required to melt the ice into water at 0 °C=55400J;
energy required to raise the temperature of this melted water to 44.6 °C =31100J;
energy supplied by the steam to change the state of 166.4 g of ice at –32.8 °C to water at 44.6 °C =97884J
Q2>What is the final mass of water in the cup at 44.6 °C??









Thanks in advance !
 
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for the first question, I tried PV=nRT ..
used the gauge pressure as P , T=288k, n=0.300998 (its the moles of gas in balloon)...
i got the answer 2.72338*10-3 .. the correct answer should be 1.93*10-3

then the second question... i used m=Q/cdT ..
I don't know which temp should I use for the change in temp...
 
I've got the first question! yey
can someone help me with the second...
seriously got no idea on that one...
 
booooo said:
then the second question... i used m=Q/cdT ..
I don't know which temp should I use for the change in temp...
Why are you using that equation then? I'm not saying you shouldn't, but if you have no idea what to plug in, it seems like you're just guessing.
 
haruspex said:
You have two sources of water, the ice and the steam. you know the mass of the ice, so put in an unknown for the mass of the steam.
What changes does each undergo along the way to arrive at the final temperature?

oh thanks! now i get the answer ;)