Question

Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.

Answer 1

Answer:

Explanation:

Using the formula below to calculate the heat absorbed in each trial:

Q = m × c × ∆T

Where;

Q = amount of heat absorbed (J)

m = mass of substance (g)

c = specific heat of water (4.184J/g°C)

∆T = change in temperature (°C)

Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.

Q = 30 × 4.184 × (40 - 0)

Q = 30 × 4.184 × 40

Q = 5,020.8J

Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.