a 100.00 mL sample of water has an initial temperature of 20.0 degrees C. The water is heated, and after 5 minutes its temperature is 50 degrees C. approximately how much energy was absorbed by the water? (the specific heat of wter is 4.18 J/g. degrees C
q = heat absorbed = mass H2O x specific heat H2O x delta T.
Q(H20)=100g x 4.18 JK-1g-1 x 30K
=100 x 4.18J x 30 (Since the rest cancel each other out)
Wallah
To calculate the amount of energy absorbed by the water, we can use the formula:
Q = m * c * ΔT
Where:
Q = Energy absorbed by the water (in joules)
m = Mass of the water (in grams)
c = Specific heat capacity of water (in J/g·°C)
ΔT = Change in temperature (in °C)
Given information:
Initial temperature (T1) = 20.0 °C
Final temperature (T2) = 50.0 °C
Mass of water = 100.00 g
Specific heat capacity of water (c) = 4.18 J/g·°C
Now, let's calculate the change in temperature:
ΔT = T2 - T1
ΔT = 50.0 °C - 20.0 °C
ΔT = 30.0 °C
Now, substitute the values in the formula to find the energy absorbed (Q):
Q = m * c * ΔT
Q = 100.00 g * 4.18 J/g·°C * 30.0 °C
Calculating the result:
Q = 12540 J
Therefore, approximately 12,540 joules of energy were absorbed by the water.