We can use the specific heat capacity formula to solve for the initial temperature:

$q = mc\Delta T$

where:

• $q = 5.19 \times 10^5$ J (heat absorbed)
• $m = 2190$ g (mass of water)
• $c = 4.184$ J/(g·°C) (specific heat capacity of water)
• $\Delta T = T_{\text{final}} - T_{\text{initial}}$

Since the water is heated until it just begins to boil, the final temperature is $100^\circ C$. We solve for $T_{\text{initial}}$:

$\Delta T = \frac{q}{mc}$

$T_{\text{initial}} = T_{\text{final}} - \Delta T$

Now, let's calculate $T_{\text{initial}}$. Given values q = 5.19e5 # Heat absorbed in J m = 2190 # Mass in g c = 4.184 # Specific heat capacity in J/(g·°C) T_final = 100 # Final temperature in °C

Calculate ΔT

delta_T = q / (m * c)

Calculate initial temperature

T_initial = T_final - delta_T T_initialThe initial temperature of the water was approximately 43.4°C.

Would you like a step-by-step breakdown of the calculation?

Related Questions:

1. What is the formula for specific heat capacity, and how is it derived?
2. How much heat is needed to completely vaporize the water after it reaches 100°C?
3. How does the specific heat capacity of water compare to other substances?
4. What happens to the temperature of water once it reaches its boiling point?
5. How would the initial temperature change if the mass of the water were doubled?

Tip: The specific heat capacity of water is relatively high, meaning it requires a lot of energy to change its temperature. This makes water an excellent substance for temperature regulation.