What is the final temperature of 0.3kg of water initially at 20 Celsius after 35 KJ Of heat is added?

Option A. 48°C

Explanation:

The following data were obtained from the question:

Mass (m) = 0.3 Kg

Initial temperature (T1) = 20°C

Heat (Q) added = 35 KJ

Specific heat capacity (C) = 4.18 KJ/Kg°C

Final temperature (T2) =..?

The final temperature of water can be obtained as follow:

Q = MC(T2 – T1)

35 = 0.3 x 4.18 (T2 – 20)

35 = 1.254 (T2 – 20)

Clear the bracket

35 = 1.254T2 – 25.08

Collect like terms

1254T2 = 35 + 25.08

1.254T2 = 60.08

Divide both side by the coefficient of T2 i.e 1.254

T2 = 60.08/1.254

T2 = 47.9 ≈ 48°C

Therefore, the final temperature of the water is 48°C.

Explanation:

Usually the angle between the y axis  and x axis is 90° and we know that for furthest travel the degree angle must be 45° with the horizontal, Mo must release the ball about halfway between straight ahead and straight up