What is the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C? What is the heat in calories? Knowing that 4.18 J = 1 calorie and C= 4.18 J/g·°C
M=25g C=4.18 J/g Tf=100 Ti=0
Q=25g x4.18j/g (100-0)
Q=25x4.18x(100)
Q=104.5(100)
Q=10,450
M=25g C=4.18 J/g Tf=100 Ti=0
Q=25g x4.18j/g (100-0)
Q=25x4.18x(100)
Q=104.5(100)
Q=10,450
M=25g C=4.18 J/g Tf=100 Ti=0
Q=25g x4.18j/g (100-0)
Q=25x4.18x(100)
Q=104.5(100)
Q=10,450
To calculate the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C, we can use the formula:
Q = m * C * (Tf - Ti)
where:
Q = heat (in Joules)
m = mass of the water (in grams)
C = specific heat capacity of water (in J/g·°C)
Tf = final temperature (in °C)
Ti = initial temperature (in °C)
Given:
m = 25 grams
C = 4.18 J/g·°C
Tf = 100 °C
Ti = 0 °C
Substituting the given values into the formula:
Q = 25g * 4.18 J/g·°C * (100 °C - 0 °C)
Q = 25 * 4.18 * 100
Q = 1,045 J
Therefore, the heat required to raise the temperature of 25 grams of water from 0 °C to 100 °C is 1,045 Joules.
To calculate the heat in calories, we can use the conversion factor 1 calorie = 4.18 J. Therefore, we can convert the heat in Joules to calories:
Heat in calories = Heat in Joules / Conversion factor
Heat in calories = 1,045 J / 4.18 J
Heat in calories = 250 calories
Therefore, the heat required to raise the temperature of 25 grams of water from 0 °C to 100 °C is 1,045 Joules or 250 calories.
To find the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C, you can use the formula Q = m * C * ΔT.
Q: heat in Joules
m: mass of the water in grams (25g)
C: specific heat of water in J/g·°C (4.18 J/g·°C)
ΔT: change in temperature (Tf - Ti)
Using the given values:
Q = 25g * 4.18 J/g·°C * (100 °C - 0 °C)
Q = 25 * 4.18 * 100
Q = 10,450 J
So, the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C is 10,450 J.
To find the heat in calories, you can use the conversion factor: 1 calorie = 4.18 J.
So, to convert from Joules to calories, divide the amount of heat in Joules by the conversion factor:
Q (in calories) = 10,450 J / 4.18 J/cal
Q (in calories) ≈ 2,500 calories
Therefore, the heat required to raise the temperature of 25 grams of water from 0 °C to 100 °C is approximately 10,450 J or 2,500 calories.