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250.0 g of 5.0°C water is heated until its temperature is 42°C.

How many joules (J) of heat energy (Q) is needed to cause this rise in temperature? _
How many calories (cal) of heat energy (Q) is needed to cause this rise in temperature? _

User Mrwooster
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1 Answer

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To calculate the amount of heat energy (Q) required to raise the temperature of the water from 5.0°C to 42°C, we can use the formula:

Q = mcΔT

Where Q is the heat energy in joules (J), m is the mass of water in grams (g), c is the specific heat capacity of water (4.184 J/g°C), and ΔT is the change in temperature in degrees Celsius (°C).

So, plugging in the values given in the problem:

m = 250.0 g

ΔT = (42°C - 5.0°C) = 37°C

c = 4.184 J/g°C

Q = (250.0 g) x (4.184 J/g°C) x (37°C)

Q = 39,050 J

Therefore, the amount of heat energy (Q) required to raise the temperature of 250.0 g of water from 5.0°C to 42°C is 39,050 J.

To convert the answer into calories (cal), we can use the conversion factor that 1 cal = 4.184 J.

So, the amount of heat energy (Q) required in calories is:

Q = (39,050 J) / (4.184 J/cal)

Q = 9,334 cal (rounded to 4 significant figures)

Therefore, the amount of heat energy (Q) required to raise the temperature of 250.0 g of water from 5.0°C to 42°C is 39,050 J or 9,334 cal.

User Krishnang K Dalal
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