An insulated beaker with negligible mass contains liquid water with a mass of 0.240 kg and a temperature of 75.8°c . how much ice at a temperature of -14.1°c must be dropped into the water so that the final temperature of the system will be 27.0°c ? take the specific heat of liquid water to be 4190 j/kg⋅k , the specific heat of ice to be 2100 j/kg⋅k , and the heat of fusion for water to be 3.34×105 j/kg .

Explanation:

Let mass of ice required is m kg

Heat gained by ice to attain zero degree

= m x 14.1 x 2100

= 29610m J

Heat gained by ice to melt = m x 3.34 x 10⁵ J

= 334000m J

Heat gained by water at zero degree to warm up to 27 degree

m x 4190 x 27 =113130 J

Total heat gained = 476740m J

Heat lost by hot water to cool up to 27 degree

.24 x 4190 x ( 75.8 - 27 )

= 49073.28 J

Heat lost = heat gained

476740 m = 49073.28

m = 49073.28/476740 kg

=102.93 gm

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

there are no contact forces mentioned on the list of choices that you've provided.