The burning of a sweet potato sample generates 102.5 kJ of heat. This heat was used to raise the temperature of 300 grams of water from its initial temperature of 15.5 degrees Celsius. Considering the specific heat of water is 4.184 J/g-degrees-C, what is the final temperature of the water

97.16°C

Explanation:

Step 1:

Data obtained from the question. This includes:

Heat (Q) = 102.5 kJ = 102.5 x 1000 = 102500J

Mass (M) = 300 g

Initial temperature (T1) = 15.5°C

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

Final temperature (T2) = ?

Change in temperatures (ΔT) = T2 - T1 = T2 - 15.5

Step 2:

Determination of the final temperature.

Applying the following equation:

Q = MCΔT

We can easily calculate the value of the final temperature as follow:

Q = MCΔT

Q = MC (T2 - T1)

102500 = 300 x 4.184 (T2 - 15.5)

102500 = 1255.2 x (T2 - 15.5)

Divide both side by 1255.2

(T2 - 15.5) = 102500/1255.2

T2 - 15.5 = 81.66

Collect like terms

T2 = 81.66 + 15.5

T2 = 97.16°C

Therefore, the final temperature of the water is 97.16°C

Answer: 97.2°c

Explanation:Given

Heat added Q=102.5KJ

Mass m=300g

Iniatial temperature t1=15.5°c

Final temperature t2=?

Specific heat capacity c=4.184

Recall Q=mc (t2-t1)

102500=300*4.184 (t2-15.5)

t2-15.5=81.66

T2=97.2°c