A plastic container with a mass of 30 grams gains 1,140 joules of heat. If the temperature of the container increases 20°C, what is the specific heat of the plastic in the container?
q = mass x specific heat x delta T
1140 = 30 x sp.h. x 20
Solve for sp.h.
To find the specific heat of the plastic in the container, we can use the formula:
q = mcΔT
Where:
q = heat gained or lost
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
In this case, we are given:
q = 1,140 joules
m = 30 grams
ΔT = 20°C
First, we need to convert the mass to kilograms by dividing it by 1000:
m = 30 grams ÷ 1000 = 0.03 kg
Now, we can rearrange the formula to solve for c:
c = q / (m * ΔT)
Plugging in the values:
c = 1,140 joules / (0.03 kg * 20°C)
Now, let's calculate the specific heat:
c = 1,140 joules / (0.03 kg * 20°C)
c ≈ 1,140 joules / 0.6 kg°C
c ≈ 1,900 joules/kg°C
Therefore, the specific heat of the plastic in the container is approximately 1,900 joules/kg°C.
To determine the specific heat of the plastic in the container, we can use the formula:
Q = mcΔT
Where:
Q = heat energy gained by the object (in joules)
m = mass of the object (in kilograms)
c = specific heat capacity of the material (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
In this case, we are given the mass of the container (30 grams), the heat gained by the container (1,140 joules), and the change in temperature (20°C).
First, we need to convert the mass from grams to kilograms:
m = 30 grams / 1000 = 0.03 kilograms
Next, we substitute the given values into the formula:
1140 joules = (0.03 kg)c(20°C)
Now we can solve for c:
c = 1140 joules / (0.03 kg × 20°C)
c = 1900 joules per kilogram per degree Celsius
Therefore, the specific heat of the plastic in the container is 1900 joules per kilogram per degree Celsius.