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Please someone help meeeeee!!!

I REALLY NEED THE ANSWERS!!!
I will be so thankful if someone answers!!!!!!!!!!!!

Please someone help meeeeee!!! I REALLY NEED THE ANSWERS!!! I will be so thankful-example-1

1 Answer

4 votes

Answer:

1 = Q = 7315 j

2 =Q = -21937.5 j

Step-by-step explanation:

Given data:

Mass of water = 50 g

Initial temperature = 20°C

Final temperature = 55°C

Energy required to change the temperature = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 j/g.°C.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 55°C - 20°C

ΔT = 35°C

Q = 50 g× 4.18 j/g.°C×35°C

Q = 7315 j

Q 2:

Given data:

Mass of metal = 100 g

Initial temperature = 1000°C

Final temperature = 25°C

Energy released = ?

Specific heat capacity = 0.225 j/g.°C

Solution:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 25°C - 1000°C

ΔT = -975°C

Now we will put the values in formula.

Q = 100 g × 0.225 j/g.°C × -975°C

Q = -21937.5 j

Negative sign show that energy is released.

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