A 4.782 kg sample of water absorbs a certain quantity of heat. The temperature increases from 17.54 ºC to 26.34 ºC. Calculate how much heat the water absorbed. Specific heat constant for water is 4.184 J/g- ºC
q = heat absorbed = mass H2O x specific heat H2O x (Ttinal-Tinitial)
Substitute the numbers and solve. Post your work if you get stuck.
not quite it. Notice that the mass you substituted is in kg but the secific heat is in units of J/GRAM*C. The easy to correct that is to change kg to grams, then recalculate.
@dr bob
Sorry.
Q = m c delta-T
= (4.782 kg)*(4.184 J/gC)*(8.80 C) = around 180 kJ
To determine the amount of heat absorbed by the water, you can use the formula:
Q = m * c * ΔT
Where:
Q = heat absorbed by the water (in Joules)
m = mass of water (in kilograms)
c = specific heat constant for water (in J/g-ºC)
ΔT = change in temperature (in ºC)
First, convert the mass of water from grams to kilograms:
m = 4.782 kg
Next, calculate the change in temperature:
ΔT = Final temperature - Initial temperature
= 26.34 ºC - 17.54 ºC
= 8.8 ºC
Now, substitute the values into the formula and solve for Q:
Q = (4.782 kg) * (4.184 J/g-ºC) * (8.8 ºC)
Let's simplify the units before calculating:
Convert the specific heat constant from J/g-ºC to J/kg-ºC:
c = (4.184 J/g-ºC) * (1 kg / 1000 g)
= 0.004184 J/kg-ºC
Now we can calculate:
Q = (4.782 kg) * (0.004184 J/kg-ºC) * (8.8 ºC)
≈ 0.1718 kg * J
Thus, the water absorbed approximately 0.1718 kilojoules (kJ) of heat.