Question

Calculate the heat energy for this scenario: You are repeating the food experiment we did in class with a different food snack. At the beginning of this experiment, you started with 25 grams of water at 22 deg * C At the end of the experiment, the final temperature of the water is 45°C. The specific heat of water is 4.18J / (deg) * C

Sub and solve

Answer 1

Mass of water (m) = 25 grams = 0.025 kg (since 1 g = 0.001 kg)

Specific heat of water (c) = 4.18 J/(g°C) = 4.18 J/(kg°C)

Initial temperature (
`T_ {initial}`) = 22°C

Final temperature (
`T_(final)` )= 45°C

Change in temperature (ΔT):

ΔT=
`T_(final)`-
`T_ {initial}`=45°−22°=23°

Now, calculate the heat energy (Q)

Q=mass×specific heat×ΔT

Q=0.025kg×4.18J/(kg°C)×23°C

Q≈2.44kJ

So, the heat energy for this scenario is approximately 2.44 kilojoules (kJ).