1. What happens to the temperature of water when 10J of energy are added if you have 10.00g of water?
2. What happens to the heat of the water if the temperature is cooled from 25.0 degrees celcius to 10.0 degrees celcius if you have 10.00g of water?
If you add 10 J energy the temperature goes up.
2. The heat of the water goes into the surroundings when the water is cooled.
Note the correct spelling of celsius.
Thank you very much
To answer these questions, we can use the equation for specific heat capacity:
Q = mcΔT
where:
Q is the heat energy transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature of the substance.
1. For the first question:
Given:
Q = 10 J (heat energy added)
m = 10.00 g (mass of water)
c = specific heat capacity of water (4.18 J/g°C, approximately)
To find ΔT (change in temperature), we rearrange the equation:
ΔT = Q / (mc)
ΔT = 10 J / (10.00 g * 4.18 J/g°C)
By performing the calculation, we find that ΔT ≈ 0.239 °C.
Therefore, when 10 J of energy are added to 10.00 g of water, the temperature of the water increases by approximately 0.239 °C.
2. For the second question:
Given:
Initial temperature (Ti) = 25.0 °C
Final temperature (Tf) = 10.0 °C
m = 10.00 g (mass of water)
c = specific heat capacity of water (4.18 J/g°C, approximately)
To find the heat energy transferred (Q), we use the equation:
Q = mcΔT
First, we need to calculate ΔT:
ΔT = Tf - Ti
ΔT = 10.0 °C - 25.0 °C
ΔT = -15.0 °C
Now, we can calculate Q:
Q = (10.00 g)(4.18 J/g°C)(-15.0 °C)
By performing the calculation, we find that Q ≈ -627 J.
Therefore, when the temperature of 10.00 g of water is cooled from 25.0 °C to 10.0 °C, the heat energy of the water decreases by approximately 627 J.
Please note that the negative sign indicates that heat energy is being removed from the water, resulting in cooling.