Answer:
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
