Question

1.Jaką ilość energii trzeba dostarczyć wodzie o masie 5 kg i temperaturze 20C, aby ogrzać ją do temperatury 100C? 2.Aby ogrzać 2 kg pewnej substancji o 1C trzeba dostarczyć jej energię 2000J. Ile wynosi ciepło właściwe tej substancji? 3.Po dostarczeniu energii równej 200kJ temperatura gliceryny zwiększyła się od 0C do 12C. Jaka była masa gliceryny, jeżeli jej ciepło właściwe to 2430 J/kg * C? Daje naj!! Potrzebne na jutro!!! 4.O ile stopni Celsjusza możemy ogrzać 2 kg wody kosztem energii 67,2 k J? Ciepło właściwe wody wynosi 4200 J/kg*0C 5.Gdy do wody o masie 1kg dostarczyliśmy 300kJ energii to jej temperatura wzrosła do 950C. Jaka była początkowa temperatura wody?

Answer 1

1) 1680 kJ

2) Specific heat capacity of substance = 1000 J/kg.°C

3) Mass of glycerol = 6.86 kg

4) 8°C

5) Initial Temperature of water = 23.57°C

Step-by-step explanation:

English Translation

1.)How much energy do you need to provide water with a weight of 5 kg and a temperature of 20C to heat it to 100C?

2. To heat 2 kg of a certain substance by 1C you need to provide its energy 2000J. What is the specific heat of this substance?

3.After supplying energy equal to 200kJ, the glycerin temperature increased from 0C to 12C. What was the mass of glycerin if its specific heat was 2430 J / kg * C? I give the best !! Needed for tomorrow !!!

4. How many degrees Celsius can we heat 2 kg of water at the expense of 67.2 k J? The specific heat of water is 4200 J/kg°.C

5. When we supplied 300kJ of energy to water with a mass of 1kg, its temperature increased to 950C. What was the initial water temperature?

Solution

The Heat required to raise the temperature of a substance from an initial temperature to a final temperature is given as

Q = mCΔT

m = mass of the substance

C = specific heat capacity

ΔT = difference in temperature from the initial to the final temperature.

1) How much energy do you need to provide water with a weight of 5 kg and a temperature of 20C to heat it to 100C?

Q = mCΔT

mass = m = 5 kg

Specific heat capacity = C = 4200 J/kg.°C (from number 4 of the question)

ΔT = 100°C - 20°C = 80°C

Q = 5 × 4200 × 80 = 1,680,000 J = 1,680 kJ

2) To heat 2 kg of a certain substance by 1C you need to provide its energy 2000J. What is the specific heat of this substance?

Q = mCΔT

Q = 2000 J

m = 2 kg

C = ?

ΔT = 1°C

2000 = 2 × C × 1

C = (2000/2) = 1000 J/kg.°C

3.After supplying energy equal to 200kJ, the glycerin temperature increased from 0C to 12C. What was the mass of glycerin if its specific heat was 2430 J / kg * C? I give the best !! Needed for tomorrow !!!

Q = mCΔT

Q = 200 kJ = 200,000 J

m = ?

C = 2430 J/kg.°C

ΔT = 12 - 0 = 12°C

200,000 = m × 2430 × 12

m = (200,000) ÷ (2430 × 12) = 6.86 kg

4. How many degrees Celsius can we heat 2 kg of water at the expense of 67.2 k J? The specific heat of water is 4200 J/kg°.C

Q = mCΔT

Q = 67.2 kJ = 67,200 J

m = 2 kg

C = 4200 J/kg°C

ΔT = ?

67,200 = 2 × 4200 × ΔT

ΔT = (67,200) ÷ (2 × 4200) = 8°C

5. When we supplied 300kJ of energy to water with a mass of 1kg, its temperature increased to 95°C. What was the initial water temperature?

Q = mCΔT

Q = 300 kJ = 300,000 J

m = 1 kg

C = 4200 J/kg.°C

ΔT = ?

300,000 = 1 × 4200 × ΔT

ΔT = (300,000)/4200 = 71.43°C

ΔT = (final temperature) - (Initial temperature)

Final temperature = 71.43°C

Initial Temperature = ?

71.43 = 95 - (Initial temperature)

Initial Temperature = 95 - 71.43 = 23.57°C

Hope this Helps!!!