An automobile cooling system holds 16 L of water. The specific heat of water is 4186 J/kg⋅C∘.How much heat does it absorb if its temperature rises from 16∘C to 94∘C?
To calculate the heat absorbed by the water in the automobile cooling system, we can use the formula:
Q = m * c * ΔT
Where:
Q = heat absorbed (in joules)
m = mass of water (in kg)
c = specific heat of water (in J/kg⋅C∘)
ΔT = change in temperature (in C∘)
We are given that the cooling system holds 16 L of water, which is equivalent to 16 kg (since 1 L = 1 kg for water). The specific heat of water is 4186 J/kg⋅C∘. And the change in temperature is from 16∘C to 94∘C.
Plugging the values into the formula:
Q = 16 kg * 4186 J/kg⋅C∘ * (94∘C - 16∘C)
Simplifying:
Q = 16 kg * 4186 J/kg⋅C∘ * 78∘C
Q = 5137152 joules
Therefore, the water in the cooling system absorbs 5,137,152 joules of heat.
To calculate the amount of heat absorbed by the water, you need to use the formula:
Q = m * c * ΔT
Where:
Q is the amount of heat absorbed (in Joules),
m is the mass of the water (in kilograms),
c is the specific heat of water (in J/kg⋅C∘),
ΔT is the change in temperature (in C∘).
To find the mass of the water, you can use the density of water, which is approximately 1 gram per milliliter, or 1000 kg per cubic meter.
Given that the automobile cooling system holds 16 L of water, you can convert it to cubic meters by dividing by 1000:
16 L = 16 * 0.001 m^3 = 0.016 m^3
Now, using the density of water, you can calculate the mass:
mass = density * volume = 1000 kg/m^3 * 0.016 m^3 = 16 kg
Next, you can substitute the values into the formula:
ΔT = 94∘C - 16∘C = 78∘C
Q = 16 kg * 4186 J/kg⋅C∘ * 78∘C = 51,747,552 J
Therefore, the amount of heat absorbed by the water is approximately 51,747,552 Joules.
Use
ΔQ=mcΔT
where
Q=heat in joules
m=mass in kg
T=temperature in °C
c=specific heat in J/(kg-°C)