Question

Someone pls help me ::/:/

How much energy must be added to 250.0 g of ice at 0.00 °C to raise the
temperature of the water to 50.0 °C? The heat of fusion for water is 334
J/g. The specific heat of water is 4.184 J/g.K.

Answer 1

Answer:

136 KJ

Explanation:

We'll begin by calculating the heat required to melt the ice. This can be obtained as follow:

Mass (M) = 250 g

Heat of fusion (Hբ) = 334 J/gºC

Heat (Q₁) required to melt the ice =?

Q₁ = MHբ

Q₁ = 250 × 334

Q₁ = 83500 J

Next, we shall determine the heat required to change the temperature. This can be obtained as follow:

Mass (M) = 250 g

Initial temperature (T₁) = 0 °C

Final temperature (T₂) = 50 °C

Specific heat capacity (C) = 4.184 J/gºC

Heat (Q₂) required to change the temperature =?

Q₂ = MC(T₂ – T₁)

Q₂ = 250 × 4.184 (50 – 0)

Q₂ = 1046 × 50

Q₂ = 52300 J

Next, we shall determine the total heat required.

Heat (Q₁) required to melt the ice = 83500 J

Heat (Q₂) required to change the temperature = 52300 J

Total heat (Qₜ) required =?

Qₜ = Q₁ + Q₂

Qₜ = 83500 + 52300

Qₜ = 135800 J

Finally, we shall convert 135800 J to KJ. This can be obtained as follow:

1000 J = 1 KJ

Therefore,

135800 J = 135800 J × 1 KJ / 1000 J

135800 J ≈ 136 KJ

Thus, the total heat required is 136 KJ.