How much heat is required to warm 122 g of water by 23.0 degrees Celsius?

q = m*c*delta T

11729.08 is the exact answer but using significant figures it would be 11700.00

lasagna

2806

Assuming that radiation with is used, that all the energy is converted to heat, and that 4.184 is needed to raise the temperature of 1.00 of water by 1.00 , how many photons are necessary to raise the temperature of a 300 cup of water from 20 to 90?

better

then

spagooter

To get the amount of heat required to warm a substance, you can use the formula:

Q = m * c * ΔT

Where:
Q is the heat required (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).

For water, the specific heat capacity (c) is approximately 4.18 J/g°C.

Now let's calculate the heat required to warm 122 grams of water by 23.0 degrees Celsius:

Q = 122 g * 4.18 J/g°C * 23.0°C

Calculating this:

Q ≈ 11,979.16 J

Therefore, approximately 11,979.16 joules of heat are required to warm 122 grams of water by 23.0 degrees Celsius.