1. An unknown amount of water was heated with 3.5 kJ, raising
its temperature from 26°C to
66°C. What was the mass of
the water?

20.93 g

Explanation:

From the question given above, the following data were obtained:

Heat (Q) = 3.5 KJ

Initial temperature (T₁) = 26°C

Final temperature (T₂) = 66°C

Mass (M) =?

Next, we shall convert 3.5 KJ to J. This can be obtained as follow:

1 KJ = 1000 J

Therefore,

3.5 KJ = 3.5 KJ × 1000 J / 1 KJ

3.5 KJ = 3500 J

Next, we shall determine the change in the temperature of the water. This is illustrated:

Initial temperature (T₁) = 26°C

Final temperature (T₂) = 66°C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 66 – 26

ΔT = 40 °C

Finally, we shall determine the mass of the water. This can be obtained as follow:

Heat (Q) = 3500 J

Change in temperature (ΔT) = 40 °C

Specific heat capacity (C) = 4.18 J/gºC

Mass (M) =?

Q = MCΔT

3500 = M × 4.18 × 40

3500 = M × 167.2

Divide both side by 167.2

M = 3500 / 167.2

M = 20.93 g

Therefore, the mass of the water is 20.93 g

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