How much heat in kcal must be added to 0.77 kg of water at room temperature (20°C) to raise its temperature to 50°C?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
mass H2O = 1000 g
specific heat H2O = 1 cal/g*c
Tfinal = 50
Tinitial = 20
To determine how much heat needs to be added to raise the temperature of water from 20°C to 50°C, we can use the specific heat formula:
Q = m * c * ΔT
Where:
Q = Heat energy (in kcal)
m = Mass of water (in kg)
c = Specific heat capacity of water (in kcal/kg°C)
ΔT = Change in temperature (in °C)
Given:
m = 0.77 kg
c = 1 kcal/kg°C (specific heat capacity of water)
ΔT = (50°C - 20°C) = 30°C
Substituting these values into the formula:
Q = 0.77 kg * 1 kcal/kg°C * 30°C
Q = 23.1 kcal
Therefore, 23.1 kcal of heat must be added to raise the temperature of 0.77 kg of water from 20°C to 50°C.
To calculate the amount of heat required to raise the temperature of water from one point to another, we can use the formula:
Q = m * c * ΔT
Where:
Q = Heat energy (in calories or kilocalories)
m = Mass of the substance (in kg)
c = Specific heat capacity of the substance (in kcal/kg°C)
ΔT = Change in temperature (in °C)
In this case, we have:
m = 0.77 kg (mass of water)
c = 1 kcal/kg°C (specific heat capacity of water)
ΔT = 50°C - 20°C = 30°C (change in temperature)
Let's substitute these values into the formula and calculate the heat:
Q = 0.77 kg * 1 kcal/kg°C * 30°C
Q = 23.1 kcal
Therefore, approximately 23.1 kcal of heat must be added to 0.77 kg of water to raise its temperature from 20°C to 50°C.