how much heat, in kilojoules, must be added to 178 grams of water to increase the temperature of the water by 50 degrees celisus?
q = mass water x specific heat water x (Tfinal-Tinitial). change to kJ.
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To calculate the amount of heat required to increase the temperature of water, we can use the equation:
Q = mcΔT
Where:
Q = amount of heat energy (in Joules)
m = mass of the water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (in Celsius)
First, we need to convert the mass of water from grams to kilograms:
Mass of water = 178 grams = 178/1000 = 0.178 kg
Now, we can calculate the amount of heat energy required:
Q = (0.178 kg) * (4.18 J/g°C) * (50°C)
Q = 37.204 Joules
To convert the amount of heat energy to kilojoules:
Q = 37.204 Joules = 37.204/1000 = 0.0372 kilojoules
Therefore, approximately 0.0372 kilojoules of heat must be added to the 178 grams of water to increase its temperature by 50 degrees Celsius.
To calculate the amount of heat required to increase the temperature of water, you can use the specific heat capacity formula:
Q = mcΔT
Where:
Q is the heat energy absorbed or released (in joules),
m is the mass of the water (in kilograms),
c is the specific heat capacity of water (in joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
First, convert the mass of water from grams to kilograms by dividing by 1000:
m = 178 grams / 1000 = 0.178 kg
The specific heat capacity of water is approximately 4.18 J/g°C. So, the conversion factor is:
1 kJ = 1000 J,
Now, calculate the heat energy required:
Q = (0.178 kg) x (4.18 J/g°C) x (50 °C)
Q = 37.3392 J
Converting from joules to kilojoules, divide by 1000:
Q = 37.3392 J / 1000 = 0.03734 kJ
Therefore, approximately 0.03734 kilojoules of heat must be added to increase the temperature of 178 grams of water by 50 degrees Celsius.