Respuesta :

Answer:

65 °C

Explanation:

Step 1: Given data

  • Mass of water (m): 5.0 g
  • Initial temperature (Ti): 25 °C
  • Heat added (Q): 832 J
  • Specific heat of water (c): 4.184 J/g.°C

Step 2: Calculate the final temperature (T)

We will use the following expression.

Q = c × m × ΔT

T = Q / c × m + Ti

T = 832 J / (4.184 J/g.°C) × 5.0 g + 25 °C

T = 65 °C