Final answer:
The amount of heat absorbed by water when a 10.0 g piece of metal at 50.0 °C is transferred into 105.0 g of water at 20.0 °C, with the water ending at 25.5 °C, is approximately 2400 Joules. The correct answer is option 4.
Step-by-step explanation:
To calculate the amount of heat absorbed by water when a 10.0 g piece of metal at 50.0 °C is transferred to a calorimeter with 105.0 g of H2O at 20.0 °C, resulting in a final temperature of the water of 25.5 °C, we'll use the formula q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat, and ΔT is the change in temperature.
The specific heat of water (CH2O) is given as 4.184 J/g°C. We use the mass of water and the change in temperature to find the heat absorbed:
qwater = mwater * CH2O * ΔTwater
qwater = (105.0 g) * (4.184 J/g°C) * (25.5 °C - 20.0 °C)
qwater = (105.0 g) * (4.184 J/g°C) * (5.5 °C)
qwater = 2406.36 J
The heat absorbed by the water is approximately 2400 J, which means that the correct answer is 2400 J.