How much heat (kJ) is absorbed by 948.0 g of water in order for the temperature to increase from 25.00cC to 32.50oC?
To calculate the amount of heat absorbed by water, you can use the equation:
q = m * c * ΔT
Where:
q is the amount of heat (in kJ),
m is the mass of the water (in grams),
c is the specific heat capacity of water (4.18 J/g°C),
ΔT is the change in temperature (in °C).
First, let's convert the mass of water from grams to kilograms:
m = 948.0 g = 0.948 kg
Next, let's calculate the change in temperature:
ΔT = (32.50°C - 25.00°C) = 7.50°C
Now, plugging these values into the equation, we can find the amount of heat absorbed by the water:
q = (0.948 kg) * (4.18 J/g°C) * (7.50°C)
q = 31.28 kJ
Therefore, the amount of heat absorbed by 948.0 g of water is 31.28 kJ.
To calculate the amount of heat absorbed by water, we need to use the heat capacity formula:
Q = m * C * ΔT
Where:
Q is the amount of heat absorbed (in kJ)
m is the mass of the substance (in grams)
C is the specific heat capacity of the substance (in J/g·°C)
ΔT is the change in temperature (in °C)
In this case:
m = 948.0 g (mass of water)
ΔT = 32.50°C - 25.00°C = 7.50°C (change in temperature)
C = 4.18 J/g·°C (specific heat capacity of water)
First, let's convert the mass to kilograms:
m = 948.0 g = 0.948 kg
Next, let's convert the specific heat capacity to kJ/kg·°C:
C = 4.18 J/g·°C = 4.18 kJ/kg·°C
Now we can calculate the amount of heat absorbed by the water:
Q = m * C * ΔT
Q = 0.948 kg * 4.18 kJ/kg·°C * 7.50°C
Q = 28.9752 kJ
Therefore, the amount of heat absorbed by 948.0 g of water to increase the temperature from 25.00°C to 32.50°C is approximately 28.98 kJ.