If 10 g of ice at - 10° C is added to 50 g of water at 15° C, what is the temperature of the mixture?
You need to understand 3 concepts to solve this answer.Firstly, You have to use the principle of calorimetry (measuring heat content of a system) to get the answer for this question.The principle goes like this:heat gained by a system(ice) = heat given by a system (water).That means the heat absorbed by ice will be equal to the heat given by the water.Secondly,now since you know the principle, next step is to know the formula of heat given or taken, that isQ= m*c*(t2-t1)where,m=mass of the substancec=heat capacity of substance (ice=2.1kJ/kgK, water=4.2kJ/kgK)t2-t1=change in temperature of the substanceNow the last but not the least concept, you must know what is latent heat, you see while there is a change of state that is when solid converts to liquid or liquid to water there is no change in temperature with the absorption of heat and that value for ice to water is 336kJ/kg.So this is the process is which is gonna happen, as the ice comes in contact with water, its temperature will rise upto 0 deg celsius. At that temperature which is the melting point of ice, it converts into water without any change in temperature. After complete conversion it again absorbs heat to reach a state where the converted ice and the water are at same temperatures (thermal equillibrium).So your final formula would be:Heat given by water = heat absorbed by ice + latent heat + heat absorbed by water (melted ice)mw*cw*(tw-T) = mice*cice*(tice-0) +336 +mmw*cmw*(T-0)The only unknown here is T =mixture temperature.Hope you got the science. :)
If we drop different weights from same height which will fall first to ground?
The answer of this question depends on the conditions of the medium that we’re performing this experiment in.In free space/vacuum, both the bodies will reach the ground at the same time. This is because the only acceleration present is earth���s gravity, which points downwards (or radially inwards towards the center of the earth, but you get the point).The time needed to reach the surface is the same for both the bodies, irrespective of their masses/weights.[math]s=ut+\frac {1} {2}at^{2}[/math]where [math]s[/math] is the distance covered, [math]u[/math] is the initial speed (which we consider zero) and [math]a[/math] is the acceleration (which we take as [math]g~[/math][math]=-9.8~m/s^2[/math]), and [math]t[/math] is the time needed.Video proof: (jump to 2:52 for the required footage)The answer to this question changes if we drop the bodies through a medium (air/liquid). Then the equation of motion is not so trivial. One then needs to consider the viscosity of the medium, the aerodynamics of the body, the weight of the body, and stuff like that. Then one cannot say with an absolute certainty (without calculating the equations of motion) that the bodies will reach the surface at the same time.Hope that helped.
A piece of cheese with a mass of 1.11kg is placed on a vertical spring of negligible mass and a force constant?
I could be facetious and say that the cheese doesn't rise at all. A giant rat, living nearby, sees the cheese on the spring and rushes forwards to gobble it all up before it goes anywhere. PS He sends you his thanks for a fine meal. But to be sensible. The energy in the spring is 1/2 k x^2 where x is 0.126 m and this causes the cheese to gain kinetic energy and subsequently potential energy. So AS MEASURED FROM THE COMPRESSED POSITION 1/2 k x^2 = mgh h = 1/2 k x^2 /(mg) = 1/2 * 2500 * 0.126^2 /(1.11 * 9.8) = 1.82 m from its release position.
If we drop a metal ball and a wooden ball from same height at the same time, then which ball will reach the ground first?
The answer to this question will vary based on the situation. Let me elaborate it situation wiseExperiment conducted in total vacuum : In this case both the metal ball and the wooden ball will hit the ground at the same time!!!, irrespective of their shape, weights and sizes. The time required to hit the ground, depends on the acceleration of the object, which in this case if ‘g’ (gravity) acting same on both the objects.Experiment conducted in atmosphere: Here we have air which fills in the atmosphere. Air being a fluid, would enable two things - Buoyancy and air-resistance. In this case, the outcome of the experiment would depend on the following parametersShape of the balls - The ball with a more aerodynamic shape would touch the ground first. As it would have lesser air-resistance, thus the resultant acceleration acting on it would be more.Volume of the ball - The ball having lesser volume will hit the ground first. Since air is a fluid, buoyancy would act on it. Buoyancy is directly proportional to the volume ( ρ x V x g).Weight of the ball - The ball having higher weight will hit the ground first. This is because of the resistive force would have lesser impact on the net acceleration (g-a).In theory, as per Galileo, the answer is indeed straight forward! But life is not that straight forward.I hope this helps!