Calculate the quantity of heat in calories absorbed when 131.01 g of water is heated from 21.1°C to 50.4°C. Enter your answer as a whole number (no decimal portion). Do not include units with your answer.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
To calculate the quantity of heat absorbed by the water, we can use the formula:
Q = m * c * ΔT
Where:
Q = Quantity of heat absorbed or released (in calories)
m = Mass of water (in grams)
c = Specific heat capacity of water (1 calorie/gram °C)
ΔT = Change in temperature (final temperature - initial temperature)
Given:
m = 131.01 g
c = 1 calorie/gram °C
Initial temperature = 21.1°C
Final temperature = 50.4°C
ΔT = (50.4°C - 21.1°C) = 29.3°C
Substituting the given values into the formula:
Q = 131.01 g * 1 calorie/gram °C * 29.3°C
Q = 3840 calories
Therefore, the quantity of heat absorbed by the water is 3840 calories.
To calculate the quantity of heat absorbed, we can use the formula:
Q = m * C * ΔT
Where:
Q is the quantity of heat in calories
m is the mass of the substance in grams
C is the specific heat capacity of the substance
ΔT is the change in temperature
For water, the specific heat capacity (C) is approximately 1 calorie/gram °C.
Given:
Mass of water (m) = 131.01 g
Initial temperature (T1) = 21.1°C
Final temperature (T2) = 50.4°C
First, let's calculate the change in temperature (ΔT):
ΔT = T2 - T1
ΔT = 50.4°C - 21.1°C
ΔT = 29.3°C
Now, let's calculate the quantity of heat absorbed using the formula:
Q = m * C * ΔT
Q = 131.01 g * 1 calorie/gram °C * 29.3°C
Calculating the above expression, we find:
Q ≈ 3849
Therefore, the quantity of heat absorbed by the water is approximately 3849 calories.