How much energy must be absorbed by 20.0g of water to increase its temp. from 283.0 C to 303.0 C?
please i really need the answer step by step correctly thanks
To calculate the amount of energy required to raise the temperature of a substance, we can use the formula:
q = m * c * ΔT
Where:
q is the amount of energy transferred (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C), and
ΔT is the change in temperature (in °C).
Step 1: Convert the given mass from grams to kilograms.
The mass is given as 20.0 grams. We need to convert this to kilograms by dividing by 1000.
Mass (m) = 20.0 g ÷ 1000 = 0.02 kg
Step 2: Determine the specific heat capacity of water.
The specific heat capacity of water is 4.18 J/g°C. However, since we are using the mass in kilograms, we need to adjust the specific heat capacity to J/kg°C.
Specific Heat Capacity (c) = 4.18 J/g°C ÷ 1000 = 4.18 J/kg°C
Step 3: Calculate the change in temperature.
The change in temperature is given as 303.0°C - 283.0°C.
ΔT = 303.0°C - 283.0°C = 20.0°C
Step 4: Substitute the values into the formula and solve for q.
q = 0.02 kg * 4.18 J/kg°C * 20.0°C
q = 1.672 J/°C * 20.0°C
q = 33.44 J
Therefore, the amount of energy that must be absorbed by the 20.0 grams of water to increase its temperature from 283.0°C to 303.0°C is 33.44 Joules.
214
i have found it
Here is how you do it.
q = mass x specific heat x (Tfinal-Tinitial).
Substitute and solve for q. You can plug in the numbers and crunch the calculator.