Answer :

Lampa20
First we need to write down heat capacity for water which is constant.
cp=4186 J/(kg*K)

The equation for Energy that we will be calculating is:

E=cp*m*T
where m is mass and T is absolute temperature (273,15 + 60 in this case). Replacing all the values in equation we get:

E = 4186*100*333,15 = 139 456 590 J
Pipperoo

Method:

To do this problem, we will calculate these three quantities:

  • Q₁: Heat needed to raise ice from -273°C to its melting point at 0°C
  • Q₂: Heat needed to melt the ice into water
  • Q₃: Heat needed to raise that water to 60°C

After we know these quantities, we will add them all together to find the total heat contained in the water.

We will use these two formulas:

Q=mcΔt

  • Q is heat
  • c is the specific heat of the substance
  • m is mass
  • Δt is the change in heat

Q=m*Lf

  • Q is heat
  • m is mass
  • Lf is the latent heat of fusion of the substance

Part 1:

To find the heat needed to raise ice from -273°C to its melting point at 0°C, we will use Q=mcΔt.

Q₁=?

m=100kg

c= specific heat of ice = 0.5kcal/kg°C

Δt= final heat - initial heat = (0°C) - (-273°C) = 273°C

Q₁=(100)(0.5)(273)

Q₁=13650kcal

Part 2:

To find the heat needed to melt the ice into water, we will use Q=m*Lf.

Q₂=?

m=100kg

Lf= latent heat of fusion of ice = 79.7kcal/kg

Q₂=(100)(79.7)

Q₂=7970kcal

Part 3:

To find the heat needed to raise the water to 60°C, we will once again use Q=mcΔt.

Q₃=?

m=100kg

c= specific heat of water (not ice anymore!) = 1.00kcal/kg°C

Δt= final heat - initial heat = (60°C) - (0°C) = 60°C

Q₃=(100)(1.00)(60)

Q₃=6000kcal

Part 4:

Now that we finally have Q₁, Q₂, and Q₃, we will add them all together to find the total heat contained in the water.

Recall that:

Q₁=13650kcal

Q₂=7970kcal

Q₃=6000kcal

Q₁+Q₂+Q₃=13650+7970+6000

Total Q = 27620kcal

In scientific notation with three significant figures:

Q = 2.77 x 10⁴ kcal

I hope this helps! Let me know if you have any questions.

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