Question

If 2,035 J of heat is added to a 500.0 g sample of water at 35.0°C, what is the final temperature of the water? Specific heat of water is 4.18 J/g°C. (Find the temperature change then subtract the initial temperature given).

Answer 1

We can start by using the equation:

Q = mCΔT

where Q is the heat added, m is the mass of the water, C is the specific heat of water, and ΔT is the change in temperature.

Plugging in the given values, we get:

2035 J = 500.0 g x 4.18 J/g°C x ΔT

Simplifying and solving for ΔT, we get:

ΔT = 2035 J / (500.0 g x 4.18 J/g°C) = 0.97°C

This is the change in temperature of the water. To find the final temperature, we need to subtract this from the initial temperature of 35.0°C:

Final temperature = 35.0°C - 0.97°C = 34.03°C

Therefore, the final temperature of the water is 34.03°C.