9. If 1,785 cal of heat is added to a 343 g sample of water (specific heat = 4.18J/g °C) at 1202, what is the *final temperature of the water? *
(1 Point)
15.76°C
031'c
18.24°C
1.24°C

17.21°C

Explanation:

Using the following formula;

Q = m × c × ∆T

Where;

Q = amount of heat in joules

m = mass of substance in grams

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

∆T = final temperature - initial temperature

According to the information provided in this question,

Q = 1,785 cal

1 cal = 4.184 J

1785 cal = 7468.4J

m = 343grams

c = 4.18J/g °C

initial temperature = 12°C

Using Q = m × c × ∆T

7468.4 = 343 × 4.18 × (T - 12)

7468.4 = 1433.74(T - 12)

7468.4 = 1433.74T - 17204.88

7468.4 + 17204.88 = 1433.74T

24673.28 = 1433.74T

T = 24673.28 ÷ 1433.74

T = 17.21°C

more infor would

a lot

c. orbital shape i hope this is correct