A 5.00-g sample of liquid water at 25.0 °C is heated by the addition of 84.0 J of energy. The final temperature of the water is __________ °C. The specific heat capacity of liquid water is 4.18 J/g-K.
How do i do this problem. Ik the answer is 29.0 .
+561 You're going to need the formula for thermal energy:
\(Q=mc\Delta T\)
Where Q is thermal energy, m is mass, c is the specific heat capacity, and delta T is change in temperature.
SImply plug the variables you were given in question as shown (make sure they're all in the correct units!):
\(84=5\times4.18\times\Delta T\)
\(\Delta T=+4.02°C\)
Add this on to your original temperature, and you get 29°C.
+561 You're going to need the formula for thermal energy:
\(Q=mc\Delta T\)
Where Q is thermal energy, m is mass, c is the specific heat capacity, and delta T is change in temperature.
SImply plug the variables you were given in question as shown (make sure they're all in the correct units!):
\(84=5\times4.18\times\Delta T\)
\(\Delta T=+4.02°C\)
Add this on to your original temperature, and you get 29°C.
