Question

To what temperature will 7900 J of heat raise 3.5 kg of water that is initially at 20.0 ∘ C ? The specific heat of water is 4186 J/kg⋅C ∘ Express your answer using three significant figures. X Incorrect; Try Again; 3 attempts remaining

Answer 1

The temperature of the water will be 20.5°C.

Specific heat capacity of water = 4186 J/ kg °C.

The temperature change can be calculated using the formula given below:

Q = mcΔT

Where

Q - heat energy

m - mass of water

c - specific heat capacity of water

ΔT - temperature change

Given:

original temperature = 20 °C

c = 4186 J/ kg °C

m = 3.5 kg

Q = 7900 J

Substituting the given values in the equation gives:

7900 J = (3.5 kg)( 4186 J/ kg °C)( ΔT)

7900 = 14651 ΔT

ΔT = 0.540 °C

Final temperature = initial temperature + ΔT

= 20.0 + 0.540

= 20.540 °C

7900 J of heat will raise the temperature of 3.5 kg of water by roughly 0.540 °C. Rounding off to 3 significant figures makes the approximate final temperature of water 20.5°C.

Answer 2

The final temperature the heat applied will raise the temperature of the water is 20.5⁰ C.

How to calculate the final temperature?

The final temperature of the water is calculated by applying the following formula as shown below;

Q = mc ΔT

where;

• m is the mass of the water
• c is specific heat capacity of water
• ΔT is change in temperature of the water

The final temperature of the water is calculated as;

7900 = 3.5 x 4186 x (T - 20)

T - 20 = (7900) / (3.5 x 4186)

T - 20 = 0.5

T = 20.5⁰ C

Thus, the final temperature of the water is 20.5⁰ C.