if 200.g of water at 20 degrees celsius absorbs 41840 j of energy, what will its final temperature be?
q = mass x specific heat x (Tfinal-Tinitial)
To find the final temperature of the water, we can use the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q = Heat energy absorbed (in joules)
m = Mass of the water (in grams)
c = Specific heat capacity of water (4.184 J/g·°C)
ΔT = Change in temperature (in °C)
Given:
Mass of water (m) = 200 g
Heat energy absorbed (Q) = 41840 J
Initial temperature (T1) = 20°C
We rearrange the formula to solve for ΔT:
ΔT = Q / (m * c)
Substituting the given values:
ΔT = 41840 J / (200 g * 4.184 J/g·°C)
Now, we can calculate ΔT:
ΔT = 41840 J / (83680 J/°C)
ΔT = 0.5 °C
To find the final temperature (T2), we add the change in temperature to the initial temperature:
T2 = T1 + ΔT
T2 = 20°C + 0.5 °C
Therefore, the final temperature of the water will be 20.5°C.
To find out the final temperature of water after absorbing a certain amount of energy, we can use the equation:
Q = mcΔT
Where:
Q = heat energy absorbed or released (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g·°C)
ΔT = change in temperature (in °C)
In this case, we have:
Q = 41840 J
m = 200 g
c = specific heat capacity of water = 4.18 J/g·°C (approximately)
Now, we need to rearrange the equation to solve for ΔT:
ΔT = Q / (mc)
Plug in the known values:
ΔT = 41840 J / (200 g * 4.18 J/g·°C)
Calculating this gives us:
ΔT ≈ 50 °C
Therefore, the final temperature of the water will be its initial temperature (20 °C) + the change in temperature (50 °C):
Final temperature = 20 °C + 50 °C = 70 °C.
Thus, the final temperature of the water will be 70 degrees Celsius.