Answer:To calculate the heat lost by 115 grams of water as it cools from 71.4°C to 34.4°C, you can use the formula for heat transfer:
Q = m * C * ΔT
Where:
Q = Heat lost (in joules)
m = Mass of the substance (in grams)
C = Specific heat capacity of water (4.18 J/g°C)
ΔT = Change in temperature (in Celsius)
First, calculate the change in temperature:
ΔT = Final temperature - Initial temperature
ΔT = 34.4°C - 71.4°C = -37°C
Now, plug the values into the formula:
Q = 115 g * 4.18 J/g°C * (-37°C)
Q = 115 g * 4.18 J/g°C * (-37)
Q ≈ -18,965.9 Joules
Explanation: Since the water is cooling down, the heat is being lost, so the heat lost is approximately -18,965.9 Joules. However, it's important to note that conventionally, heat lost is considered as a positive value, so you can express it as approximately 18,965.9 Joules.