If a gram of water at sea level is heated from 37 degrees Celsius to 101 degrees. How much heat would be absorbed by the water?

q = mass x specific heat x (Tfinal-Tinitial)

4284.416. but what unit would it be in?

The mass H2O is in grams in the problem. What unit did you use for specific heat? If calories/gram then q is in calories. If in joules/gram then q is in joules.

To calculate the amount of heat absorbed by the water, you can use the specific heat capacity formula:

Q = m * c * ΔT

Where:
Q = heat absorbed or released (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in Joules/g°C)
ΔT = change in temperature (in °C)

In this case, the mass of the water is given as 1 gram, the specific heat capacity of water is approximately 4.18 Joules/gram°C, and the change in temperature is 101 - 37 = 64°C.

Plugging these values into the formula:

Q = 1g * 4.18 J/g°C * 64°C
Q = 267.52 Joules

Therefore, the water would absorb approximately 267.52 Joules of heat when heated from 37°C to 101°C.