Answer:
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
