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

3 answers

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.
are you sure bot?
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.
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