High School

If you have Q = 125,580 J of heat and apply it to some water of unknown mass that has an initial temperature of 10°C to 255°C, what is the mass of the water?
(Given: Specific heat capacity of water, c = 4,186 J/kg·K)

- 0.5 kg
- 1 kg
- 2.5 kg
- 4 kg

Answer :

The mass of the water is approximately 0.122 kg.

To find the mass of the water, we can use the formula for heat transfer:

[tex]Q = mc\Delta T[/tex]

where:

  • [tex]Q[/tex] is the heat added (125,580 J)
  • [tex]m[/tex] is the mass of the water
  • [tex]c[/tex] is the specific heat capacity of water (4,186 J/kg·K)
  • [tex]\Delta T[/tex] is the change in temperature (255°C - 10°C)

First, let's calculate the change in temperature, [tex]\Delta T[/tex]:

[tex]\Delta T = 255- 10= 245[/tex]

Now we can rearrange the heat transfer formula to solve for mass, [tex]m[/tex]:

[tex]m = \frac{Q}{c \Delta T}[/tex]

Substitute the given values into the equation:

[tex]m = \frac{125,580 J}{4,186 \frac{J}{kg·K} \times 245 K}[/tex]

Calculate the denominator:

[tex]4,186 \times 245 = 1,025,570[/tex]

Now, divide the heat by the result:

[tex]m = \frac{125,580}{1,025,570} \approx 0.122 \text{ kg}[/tex]

Therefore, the mass of the water is approximately 0.122 kg.

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