Question

Use the following information to answer the next question.
Ali places a beaker containing 500 g of water, with a thermometer inside, on a burner, and notes down the initial temperature of water as 25.0 oC. After 10.0 minutes, the temperature went up to 75.0 oC. The specific heat of water is 4.19 J/g oC.

The quantity of heat gained by the water is:
Question 12 options:

a) 10475 kJ
b) 157 kJ
c) 105 kJ
d) 523 kJ
e) 1048 kJ

Answer 1

The quantity of heat gained by the water can be calculated using the equation:

q = mcΔT

Where:
q = quantity of heat gained (in Joules)
m = mass of water (in grams)
c = specific heat of water (in J/g oC)
ΔT = change in temperature (in oC)

Given:
m = 500 g
c = 4.19 J/g oC
ΔT = 75.0 oC - 25.0 oC = 50.0 oC

Plugging these values into the equation:

q = 500 g * 4.19 J/g oC * 50.0 oC
q = 104750 J

Converting to kilojoules:

q = 104750 J / 1000 = 104.75 kJ

The quantity of heat gained by the water is 104.75 kJ.

Therefore, the correct answer is a) 10475 kJ.

Answer 2

are you sure bot?

Answer 3

Apologies for the confusion in my previous response. Let's recalculate the quantity of heat gained by the water using the correct equation:

q = mcΔT

Given:
m = 500 g
c = 4.19 J/g oC
ΔT = 75.0 oC - 25.0 oC = 50.0 oC

Plugging these values into the equation:

q = 500 g * 4.19 J/g oC * 50.0 oC
q = 104750 J

Converting to kilojoules:

q = 104750 J / 1000 = 104.75 kJ

Apologies for the confusion earlier. The correct answer is indeed a) 10475 kJ.