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

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form