QUESTION 43 How much energy does it take to heat 5.00 g of solid water from...

Question

Question

Answer 1

Answer #1

-43 ma Ti= -10°c so = 5x13 kg Ray Heat taken m sa ( 5x103). (2090 J kg-c 155) $cc). = 104.5 J option-c 9-५५ 59 Тəе Чәа oQ =

Answer 2

Similar Homework Help Questions

QUESTION 46 How much energy does it take for 5.00 g of solid water to melt?...

QUESTION 46 How much energy does it take for 5.00 g of solid water to melt? Latent heats: fusion = 3.33x105 y/kg vaporization - 2.26x106kg O A. 166.5) B. 1.66x106) C. 11,300) OD. 1,665) QUESTION 47 What is the energy change of water when 5.00 g of gaseous water condenses? Latent heats: fusion - 3.33x105 kg, vaporization - 2.26x106j/kg A. 11,300) OB. 16,650) C. -11,300 OD.-16,650)
How much energy is required to change a 37 g ice cube from ice at −13◦C...

How much energy is required to change a 37 g ice cube from ice at −13◦C to steam at 113◦C? The specific heat of ice is 2090 J/kg ·◦ C, the specific heat of water is 4186 J/kg ·◦ C, the specific heat of stream is 2010 J/kg ·◦ C, the heat of fusion is 3.33 × 105 J/kg, and the heat of vaporization is 2.26 × 106 J/kg. Answer in units of J.
How much energy is required to change a 47 g ice cube from ice at -12...

How much energy is required to change a 47 g ice cube from ice at -12 C to steam at 117°C? The specific heat of ice is 2090 J/kg. C, the splecific heat of water is 4186 J/kg C, the specific heat of stream is 2010 J/kg. C, the heat of fusion is 3.33 x 105 J/kg, and the heat of vaporization is 2.26 x 106 J/kg Answer in units of J

014 10.0 points How much energy is required to change a 35 g ice cube from...

014 10.0 points How much energy is required to change a 35 g ice cube from ice at -12°C to steam at 111°C? The specific heat of ice is 2090 J/kg.° C, the specific heat of water is 4186 J/kg.° C, the specific heat of stream is 2010 J/kg.° C, the heat of fusion is 3.33 x 10° J/kg, and the heat of vaporization is 2.26 × 106 J/kg. Answer in units of J
How much heat must be removed from 456 g of water at 25.0 degree C to...

How much heat must be removed from 456 g of water at 25.0 degree C to change it into ice at -10.0 degree C? The specific heat of ice is 2090 J/kg K. the latent heat of fusion of water is 33.5 times 10^4 J/Kg, and the specific heat of water is 4186 J/kg K.
015 10.0 points no x How much energy is required to change a 35 g is...

015 10.0 points no x How much energy is required to change a 35 g is ice cube from ice at -12C to steam at 113 C? ber The specific heat of ice is 2090 J/kg.° C, the specific heat of water is 4186 J/kg.°C, the specific heat of stream is 2010 J/kg. C, the red heat of fusion is 3.33 x 105 J/kg, and the heat 1s of vaporization is 2.26 × 100 J/kg. nd Answer in units of...

How much heat is required to change 456 g of ice at -20.0Degree C into water...

How much heat is required to change 456 g of ice at -20.0Degree C into water at 25.0Degree C? specific heat of water = 4186]/(kg-K); specific heat of ice = 2090 J/(kg.K) and latent heat of fusion of water = 33.5 times 10^4 J/kg.
How much energy is required to turn 5.00 kg of ice at -25.0 degrees C to...

How much energy is required to turn 5.00 kg of ice at -25.0 degrees C to steam at 140 degrees C? Show your work. cice = 2090 J/kg•oC Lfusion (water) = 3.33 x 105 J/kg cwater = 4186 J/kg•oC Lvapor(water) = 2.26 x 106 J/kg csteam = 2010 J/kg•oC
Question 13 (1 point) How much heat (in MJ) does it take to completely convert 0.80...

Question 13 (1 point) How much heat (in MJ) does it take to completely convert 0.80 kg of ice initially at a temperature of -15°C to steam (water vapor) at a final temperature of 130°C? Specific heats (in J/kg.K): ice (2100), water (4190), steam (2000) Latent heats (in J/kg): fusion (3.34 x 105), vaporization (2.256 X 106) 1.0 2.0 1.5 2.5 0.5 3.0

1. How much heat energy is required to change a 0.50-kg of ice at – 10...

1. How much heat energy is required to change a 0.50-kg of ice at – 10 0C to steam at 110 0C? [Lf = 3.33 x 105 J/kg, c ice = 2090, c water = 4186 J/kg.0C, Lv = 2.26 x 106 J/kg, c steam = 2010 J/kg.0C]