The amount of heat absorbed by the water is 1606.6 J (383.99 Calories )
- We'll begin by calculating the change in the temperature of the water.
Initial temperature (T₁) = 21.5 °C
Final temperature (T₂) = 28.1 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 28.1 – 21.5
ΔT = 6.6 °C
- Finally, we shall determine the heat absorbed by the water. This can be obtained as follow:
Mass of water (M) = 58.18 g
Change in temperature (ΔT) = 6.6 °C
Specific heat capacity of water (C) = 4.184 J/gºC
Heat absorbed (Q) =?
Q = MCΔT
Q = 58.18 × 4.184 × 6.6
Q = 1606.6 J
Divide by 4.184 to express in Calories
Q = 1606.6 / 4.184
Q = 383.99 Calories
Therefore, the amount of heat absorbed by the water is 1606.6 J (383.99 Calories )
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