78,088.72J of heat is added to 1000 g of ice at -10 oC. How much ice...

Question

Question

78,088.72J of heat is added to 1000 g of ice at -10 oC. How much ice melts?

science physics

Answer 1

Answer #1

78,088.72 = 1000X 2.1 X 10 + m X 334000

solve for m

Answer 2

Answer #2

Heat required to take ice to 0 oC=mcΔT=1000*2.1*10=21,000 J

Remaining heat=77079.1-21000=56079.1 J

Amount of ice that will melt= Remaining heat/Lf= 56079.1/334 =167.9 g

Answer 3

Answer #3

59468.5 = m(335000). m = 177.5g

Answer 4

Similar Homework Help Questions

59,231.41J of heat is added to 1000 g of ice at -10 oC. How much ice...

59,231.41J of heat is added to 1000 g of ice at -10 oC. How much ice melts?
59,231.41J of heat is added to 1000 g of ice at -10 oC. How much ice...

59,231.41J of heat is added to 1000 g of ice at -10 oC. How much ice melts?
91,375.73J of heat is added to 1000 g of ice at -10 oC. How much ice...

91,375.73J of heat is added to 1000 g of ice at -10 oC. How much ice melts?

47,454.25J of heat is added to 1000 g of ice at -10 oC. How much ice melts?...

47,454.25J of heat is added to 1000 g of ice at -10 oC. How much ice melts?
30,501.39J of heat is added to 1000 g of ice at -10 oC. How much ice melts?

30,501.39J of heat is added to 1000 g of ice at -10 oC. How much ice melts?
How much heat is required to change 50 g of ice at 0 oC into liquid...

How much heat is required to change 50 g of ice at 0 oC into liquid water at 0 oC? Qmelt ice = cal 2) If 300 g of water initially at 25 oC released the amount of heat that you found in Part (1), what would its temperature be afterward? Tafter = oC 3)If you were to combine 50 g of liquid water at 0 oC with 300 g of the water at the temperature you found in Part...

How much heat must be absorbed by 20 g of ice at minus 16 oC to...

How much heat must be absorbed by 20 g of ice at minus 16 oC to transform into 20 g of water at 45 oC. (cice= 2.05 kJ/kg, cwater=4.18 kJ*K/kg, csteam=2.01 kJ*K/kg; Lfwater=333.5 kJ/kg @ 0 oC, Lvwater= 2257 kJ/kg @100 oC) Q= J
A 40-g block of ice is cooled to -71°C and is then added to 610 g...

A 40-g block of ice is cooled to -71°C and is then added to 610 g of water in an 80-g copper calorimeter at a temperature of 23°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. (If not all the ice melts, determine how much ice is left.) Remember that the ice must first warm to 0°C, melt, and then continue warming as water. (The specific heat of ice is 0.500 cal/g·°C = 2,090...
A 1.65 kg block of ice at 0°C is added to a pienic cooler. How much...

A 1.65 kg block of ice at 0°C is added to a pienic cooler. How much heat will the ice remove as it melts to water at 0°C? kcal

One ice cube at -10oC weighing 30 g is added to a glass containing 200 g...

One ice cube at -10oC weighing 30 g is added to a glass containing 200 g of water at 20oC. During the ice melting, additional 3.0 kJ of the heat are transferred from the glass wall to the ice-water mixture. The latent heat of fusion of the ice is 334 kJ/kg, the specific heat of the ice is 1.93 kJ/kg oC; and the specific heat of the water is 4.18 kJ/kg oC. Calculate the final water temperature when the ice...