A liter of water is used to cool some electronics. If 1000 J of heat is given off by the electronics, by what
temperature does the water increase? One liter of water has a mass of 1 kg and a specific heat of 4.184 J/(g
°C).
I can not get this..please help
From the given information, we deduce that
It takes 4.184 J/g * 1000 g = 4184 J to raise the temperature of 1 L of water by 1°C.
Assuming that all 1000J of heat are absorbed by the 1 L of water, then by proportion, the temperature increase equals 0.239°C.
1°C * (1000/4184)
To determine the temperature increase of water when it absorbs a certain amount of heat, you can use the equation:
Q = mcΔT
Where:
Q is the heat energy absorbed or released by the substance (in Joules),
m is the mass of the substance (in kilograms),
c is the specific heat capacity of the substance (in Joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
In this case, we are given the following information:
Q = 1000 J (heat given off by the electronics)
m = 1 kg (mass of water)
c = 4.184 J/(g °C) (specific heat capacity of water)
First, we need to convert the mass of water from grams to kilograms since the specific heat capacity is provided in terms of grams:
m = 1 kg = 1000 g
Now, we can substitute the values into the equation and solve for ΔT:
1000 J = (1000 g) * (4.184 J/(g °C)) * ΔT
Dividing both sides of the equation by (1000 g) * (4.184 J/(g °C)), we get:
ΔT = (1000 J) / ((1000 g) * (4.184 J/(g °C)))
Calculating this expression, we find:
ΔT ≈ 0.238 °C
Therefore, the temperature of the water increases by approximately 0.238 degrees Celsius when it absorbs 1000 Joules of heat from the electronics.