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
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
2 answers
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
Q=25g x4.18j/g (100-0)
Q=25x4.18x(100)
Q=104.5(100)
Q=10,450