Determine how many calories are need to heat 30 grams of water from 10 degrees C to 30 degrees C
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
600 calories
To determine the number of calories needed to heat 30 grams of water from 10 degrees Celsius to 30 degrees Celsius, you can use the specific heat capacity formula.
The specific heat capacity of water is approximately 4.18 joules per gram per degree Celsius (4.18 J/g·°C).
First, let's convert the grams of water to joules:
30 grams × 4.18 J/g·°C = 125.4 joules
Next, we need to account for the temperature change from 10 degrees Celsius to 30 degrees Celsius:
Temperature change = final temperature - initial temperature
Temperature change = 30°C - 10°C = 20°C
Now, multiply the energy required per degree Celsius by the temperature change:
125.4 joules × 20°C = 2508 joules
Finally, we can convert joules to calories by dividing the value by 4.184 joules per calorie:
2508 joules / 4.184 joules/calorie = approximately 599.14 calories
Therefore, it takes approximately 599.14 calories to heat 30 grams of water from 10 degrees Celsius to 30 degrees Celsius.
To determine the number of calories needed to heat 30 grams of water from 10 degrees Celsius to 30 degrees Celsius, we can use the formula:
Q = m * c * ΔT
where:
Q is the amount of heat energy required (in calories)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in calories/gram °C)
ΔT is the change in temperature (in °C)
For water, the specific heat capacity is approximately 1 calorie/gram °C.
Let's substitute the values into the formula:
Q = 30 grams * 1 calorie/gram °C * (30 °C - 10 °C)
Q = 30 * 1 * 20 = 600 calories
Therefore, 600 calories are needed to heat 30 grams of water from 10 degrees Celsius to 30 degrees Celsius.