How do you solve this

23.6°C

Explanation:

Step 1:

Data obtained from the question. This include the following:

Mass = 0.125kg = 0.125 x 1000 = 125g

Initial temperature (T1) = 22°C

Heat (Q) generated in 4.5mins = 835J

Specific heat capacity (C) of water = 4.184J/g°C

Change in temperature (ΔT) =?

Final temperature (T2) =?

Step 2:

Determination of the change in temperature, ΔT of water.

This can be obtained as shown below:

Q = MCΔT

835 = 125 x 4.184 x ΔT

Divide both side by 125 x 4.184

ΔT = 835 / (125 x 4.184)

ΔT = 1.6°C

Therefore, the change in temperature, ΔT is 1.6°C

Step 3:

Determination of the final temperature of water.

This can be obtained as follow:

Initial temperature (T1) = 22°C

Change in temperature (ΔT) = 1.6°C

Final temperature (T2) =?

Change in temperature (ΔT) = Initial temperature – Final temperature

ΔT = T2 – T1

1.6 = T2 – 22

Collect like terms

T2 = 1.6 + 22

T2 = 23.6°C

Therefore, the final temperature of the water is 23.6°C.

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