From the statements in the question, the starting temperature before 615 calories of heat was added is 24 °C.
Given that the total energy added to the water = Heat added to raise the temperature of the water to boiling point + latent heat of vaporization of water
Mass of the water = 1 g
Boiling point of water = 100°C
Starting temperature of the water = θ°C
Latent heat of vaporization of water = 539 cal/g
Specific heat capacity of water = 1 cal/g°C
Hence;
H= mc(100 -θ) + mL
H = m(c(100 -θ) + L)
H = 1[1 × (100 - θ) + 539)]
615 = 100 - θ + 539
θ = 100 + 539 - 615
θ = 24 °C
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