If you wish to warm 120 kg of water by 16°C for your bath, how much heat is required?
((The heat capacity of liquid water is 4186 J/(kg·oC) )
Q = M C *(delta T)
delta T = 16 C (it means "change in temperature)
M = 120 kg
C = 4186 J/(kg·oC
Multiply it out for the answer (Q, the heat required), in Joules
8037120
660
To calculate the amount of heat required to warm 120 kg of water by 16°C, we can use the formula:
Q = M * C * (delta T)
where:
Q = heat required (in Joules)
M = mass of water (in kilograms)
C = specific heat capacity of water (in Joules per kilogram per degree Celsius)
delta T = change in temperature (in degrees Celsius)
In this case, we have:
M = 120 kg
C = 4186 J/(kg·°C)
delta T = 16°C
Substituting these values into the formula, we get:
Q = 120 kg * 4186 J/(kg·°C) * 16°C
Now, let's calculate it:
Q = 120 kg * 4186 J/(kg·°C) * 16°C
= 8025600 J
So, the amount of heat required to warm 120 kg of water by 16°C is 8,025,600 Joules.