Question

How much heat in joules is released when 45.0 g of water (C = 4.184 J/gºC) is

cooled from 25.0°C to 23.0°C?

Answer 1

– 376.56 J

Step-by-step explanation:

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

Mass (M) = 45 g

Specific heat capacity (C) = 4.184 J/gºC

Initial temperature (T1) = 25 °C

Final temperature (T2) = 23 °C

Heat released (Q) =..?

Next, we shall determine the change in the temperature of the water. This can be obtained as follow:

Initial temperature (T1) = 25 °C

Final temperature (T2) = 23 °C

Change in temperature (ΔT) =.?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 23 – 25

Change in temperature (ΔT) = – 2 °C

Finally, we shall determine the heat released as follow:

Mass (M) = 45 g

Specific heat capacity (C) = 4.184 J/gºC

Change in temperature (ΔT) = – 2 °C

Heat released (Q) =..?

Q = MCΔT

Q = 45 × 4.184 × (– 2)

Q = – 376.56 J

Therefore, the heat released is – 376.56 J.

Note: The negative sign indicates that heat is released to the surrounding.