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When 100 $\mathrm{mL}$ of 9 $\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ at $0^{\circ} \mathrm{C}$ is added to 100 $\mathrm{mL}$ of liquid water at $0^{\circ} \mathrm{C},$ the temperature rises to $12^{\circ} \mathrm{C} .$ When 100 $\mathrm{mL}$ of 9 $\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ at $0^{\circ} \mathrm{C}$ is added to 100 $\mathrm{g}$ of solid ice at
$0^{\circ} \mathrm{C},$ the temperature falls to $-12^{\circ} \mathrm{C}$ . Explain the difference in behavior.

Do this problem. We need to assume that the heat that is gained by glass is equivalent to the heat lost by H 20 by the water. So our problem for this is going to be mass of the glass trying to specific heat of the glass. Time to change in temperature of the glass, which is equal to since he gained by glass has to be equal to heat loss. My water, you will be equal to the mass of water times a specific heat of water, time to change in temperature of the water. So if we go ahead and plug in the information that we know into this problem, so the glass has a mass of 35 Gramps, so we'll start with 35 grams. The specific heat of glass is your point too. Put our units and then our change in temperature which is 39.2 degrees Celsius minus 21.6 degrees Celsius. Okay, so then, for the other side are massive water. We have 135 milliliters. Since leaders are equivalent to kilograms, we know that we have 1 35 grams of water under specific heat, and then we know that we went down too far. We want our original temp of water. So I'm gonna go down the line here, but continue the equation so we would multiply that times our original time or temperature. See any water minus the final temperature, which is 39.2 degrees Celsius. So if we go ahead and do the math for this to get our answer for our water temperature, we end up with 40.12 degrees.

To solve this problem, I will use the formula vita is equal to B two minus B. Even by Beaven multiplication. T two minus even. So this is the formula where the one is the initial volume between the final volume T one is initial temperature t to its final temperature. And just putting the value in this expression, I can write 1.383 liter minus 1.5 liter by yes, 15 liter 1.5 liter multiplication, 15 C minus 97.1 degrees centigrade on simplification I get the value of Byetta is equal to minus 0.117 liter by 1.5 leader multiplication and minus point minus 82 points one degrees, integrate on simplification. I get the value of Vita is equal to 950 multiplication, 10 to depart minus six part degrees pulse integrated so the value it match with the liquid in the vessel, each gasoline. So it will be guess so lean as our answer for this problem.

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Problem Number 34 you have to write what is the temperature after meeting in a constant pressure? So first that gas is in experience. The reflective nature one but temperature. What if it's wanted to on the sub's Celsius on volume off that particular temperature gas is 857. Win dated. Now it's going to be heated and expert in the constant pressure that is very important term constant pressure on the temperature to accepting temperature. We don't know it, and we do is give up. So first of all these should convert this decree substances into an absolute temperature. Just add 273 and then you get that can be number can be temporary ship. So then, since this is in constant pressure, you can use child slow, which is this one. We proportional toe temple, Asian volumes, proportions with temperature at constant friction. So when you solve this equation 42 for the inspiration for the second state, we'll get we to t one divided by everyone. So now we're going to substitute all the values on. Then when you cancels out this one, you will get five or 5 12 e, but we need it in the census. So you are goingto we got 273 to get the Celsius Valley, which is 232 ridiculous off Celsius.