Final answer:
The energy transferred when 250.0 mL of water is warmed from 25 °C to 37 °C, with a specific heat capacity of 4.16 J/(mL °C), is 12.48 kJ.
Step-by-step explanation:
The student is asking how much energy is transferred when 250.0 mL of water is warmed from 25 °C to 37 °C, given that the specific heat capacity of water is 4.16 J/(mL °C). To find the energy transferred, we use the formula Q = mcΔT, where:
Q is the heat energy transferred (in Joules)
m is the mass of the water (in grams)
c is the specific heat capacity (in J/(g°C))
ΔT is the change in temperature (°C)
Since 1 mL of water is approximately equal to 1 g, we can use the volume in mL for the mass. Thus, the energy transferred can be calculated as follows:
Q = (250.0 mL) x (4.16 J/(mL °C)) x (37 °C - 25 °C)
Q = (250.0 mL) x (4.16 J/(mL °C)) x (12 °C)
Q = 12,480 J or 12.48 kJ
This calculation shows that the chemical reaction caused a transfer of approximately 12.48 kJ of energy to heat the water from 25 °C to 37 °C.