Final answer:
The amount of heat required to increase the temperature of 90 g of water from 0°C to 100°C is 37,656 Joules or 37.656 kJ, using the specific heat capacity of water which is 4.184 J/g°C.
Step-by-step explanation:
The student is asking about the amount of heat required to raise the temperature of a certain mass of water from one temperature to another. To calculate this, we use the equation ΔQ = mcΔT, where ΔQ is the heat added, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. Given that the specific heat capacity for water is actually 4.184 J/g°C and not 1 cal/(g°C) as initially stated, we can perform the calculation.
First, we must convert the mass of water from grams to kilograms by dividing by 1000 since we are using the specific heat capacity in J/kg°C. Next, we can calculate the amount of heat required to raise the temperature of 90 g (0.09 kg) of water from 0°C to 100°C:
Heat required (ΔQ) = mcΔT
= (0.09 kg)(4,184 J/kg°C)(100°C - 0°C)
= 0.09 kg × 4,184 J/kg°C × 100°C
= 37,656 J
Therefore, the heat required to raise the temperature of 90 g of water from 0°C to 100°C is 37,656 Joules or 37.656 kJ.