Final answer:
To heat 50 mL of water from 0°C to 60°C, the energy used is 3000 calories, which can be converted to 12.6 kJ.
Step-by-step explanation:
The student wants to calculate the energy used to heat 50 mL of water from 0°C to 60°C.
In order to find the energy consumed, we use the calorimetry formula q = mcΔT, where q is the heat energy, m is the mass in grams, c is the specific heat capacity, and ΔT is the change in temperature.
The specific heat capacity of water is 1 cal/g°C. Given that the density of water is 1 g/mL, 50 mL of water equals 50 grams. Therefore, we can calculate the energy as follows:
q = mΔT = 50g × (60°C - 0°C)
= 50g × 60°C
= 3000 calories or 12.6 kJ (since 1 cal = 4.184 J)