Iced tea is made by adding ice to 1.8 kg of hot tea. the hot tea is initially at 80˚c. 1.4 kg of ice, initially at –10˚c, is placed in the tea and allowed to come to thermal equilibrium inside a thermos. you can assume the hot tea is essentially all water. what will be the final temperature of this system?

Explanation:

Given:

We, know

Heat required for warming of ice upto 0°C from -10°C

Heat released during the cooling of tea upto 0°C from 80°C

∵

∴The heat from the quantity that remains after bringing the ice to 0°C will be used for melting the ice.

Heat remaining afterthis process:

= 572984 J

Amount of heat energy required by the total mass of ice to melt at 0°C:

Now, the remaining of the heat will be used to raise the temperature of the tea (water) from 0°C to a higher temperature.

Heat remaining after melting the total ice into water at 0°C which will increases the temperature of water:

Now, the final temperature of the tea(water)rising from 0°C:

∵whole ice(along with tea) is now water with initial temperature = 0°C

where:

= final temperature

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft