You have a 0.50kg of ice at -200C and it is placed in a cup containing 3.5kg of water that is at 250C. The cup is highly insulated so no heat flows into or out of the cup. A) Determine if all the ice melts. B) What is the final temperature of the water in the cup

a) All ice melts

b) Final temperature =  10.6°C

Explanation:

Given Data:

mass of ice = 0.50 kg

temperature = -20°C

placed in a cup containing 3.5kg of water that is at 25°c

A) Determine if all the ice melts

3.5 kg of 25°c  water to 0°c

Heat rejected = 3.5 * 10^3 * 1 * 25 cal = 87500 cal

0.5 kg - 20°c ice to 0°c ice

Heat gained = 0.5 * 0.5 * 10^3 *20 cal = 5000 cal

0.50 kg 0°c ice to 0°c water

Heat gained = 0.5 * 10^3 * 80 cal  = 40000 cal

All the ice will melt eventually

B) Determine the final temperature

0°c water of 0.5kg to t°C water

heat gained = 0.5 * 10^3 * 1 * t = 500 t cal

heat rejected = 3.5*10^3 * 1 * ( 25 - t )

                      = 3500 ( 25 - t )

To determine the Final temperature we will equate the Total heat gained to the Total heat rejected

= 5000 + 40000 + 500 t = 3500 ( 25 - t )

= 4000 t = 3500 * (25 - 45000)

Hence t = 10.6°C

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: