If 20.0 kilocalories of energy were added to 80.0 g of water, what change in temperature would be observed? Please explain how.

250 °C.

Explanation:

From the question given above, the following data were:

Heat (Q) required = 20.0 kilocal

Mass (M) of water = 80 g

Change in temperature (ΔT) =.?

Next, we shall convert 20.0 Kilocalories to calories. This can be obtained as shown below:

1 kilocal = 1000 Cal

Therefore,

20 kilocal = 20 kilocal × 1000 Cal / 1 kilocal

20 kilocal = 20000 calories

Therefore, 20 kilocalorie is equivalent to 20000 calories.

Finally, we shall determine the change in temperature observed as follow:

Heat (Q) required = 20000 Cal

Mass (M) of water = 80 g

Specific heat capacity (C) of water = 1 Cal /gºC

Change in temperature (ΔT) =.?

Q = MCΔT

20000 = 80 × 1 × ΔT

20000 = 80 × ΔT

Divide both side by 80

ΔT = 20000 / 80

ΔT = 250 °C

Therefore, the change in temperature observed is 250 °C