Physics, 17.07.2020 14:01 katieP1992

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

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

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Answer from: cjgonzalez981

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.

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

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