Ethylene glycol has a specific heat of 0.578 cal/(g×°C). If 23.2 g of ethylene glycol absorbs 75.6 cal of heat energy, what will the temperature increase be?
Well, if there's one thing I know about ethylene glycol, it's that it's not one to show off with its heat capacity. But let's crunch some numbers here.
Given that ethylene glycol has a specific heat of 0.578 cal/(g×°C), and you have 23.2 g of it, we can multiply those two values together to find out its total heat capacity.
So, 23.2 g × 0.578 cal/(g×°C) equals 13.3856 cal/°C. That's the heat capacity of your ethylene glycol.
Now, since you said that it absorbs 75.6 cal of heat energy, we can divide that number by the heat capacity to find the temperature increase.
75.6 cal ÷ 13.3856 cal/°C equals approximately 5.652°C.
Therefore, the temperature of your ethylene glycol will increase by approximately 5.652 degrees Celsius. So, don't be surprised if the ethylene glycol starts feeling a bit hot under the collar!
To calculate the temperature increase, we can use the formula:
Q = mcΔT
Where:
Q is the heat energy absorbed (in calories)
m is the mass of the substance (in grams)
c is the specific heat (in cal/(g*°C))
ΔT is the change in temperature (in °C)
In this case, we are given:
Q = 75.6 cal
m = 23.2 g
c = 0.578 cal/(g*°C)
Rearranging the formula, we have:
ΔT = Q / (mc)
Substituting the given values:
ΔT = 75.6 cal / (23.2 g * 0.578 cal/(g*°C))
ΔT = 75.6 cal / (13.4096 cal/°C)
ΔT ≈ 5.64 °C
Therefore, the temperature increase will be approximately 5.64 °C.
To calculate the temperature increase, we can use the equation:
ΔT = Q / (m × C)
Where:
ΔT is the temperature change
Q is the amount of heat energy absorbed
m is the mass of the substance
C is the specific heat capacity
Given:
Q = 75.6 cal
m = 23.2 g
C = 0.578 cal/(g×°C)
Substituting the values into the equation:
ΔT = 75.6 cal / (23.2 g × 0.578 cal/(g×°C))
First, let's simplify the units of cal on the right side of the equation:
ΔT = 75.6 / (23.2 × 0.578 (g×°C))
Now, multiply the denominator:
ΔT = 75.6 / 13.3856 (g×°C)
Finally, divide the numerator by the denominator:
ΔT ≈ 5.65 °C
Therefore, the temperature will increase by approximately 5.65 °C.
q = mass x specific heat x (Tfinal-Tinitial)
75.6 = 23.3 x 0.578 x delta T.
Solve for delta T.