Question

An iron pin (specific heat=0.473 J/g degrees Celsius) at 110 degrees Celsius is dropped into 50.0 mL of H2O at 15.0 degrees Celsius. After a couple minutes, the temperature of the water reaches 17.4 degrees Celsius. What was the mass of the iron pin?

Answer 1

The mass of the iron pin is calculated using the conservation of energy principle by equating the heat lost by the pin to the heat gained by the water, resulting in a mass of approximately 114.73 grams.

Step-by-step explanation:

To find the mass of the iron pin, we will use the concept of conservation of energy, where the heat lost by the iron pin will equal the heat gained by the water. The formula for heat exchange is q = mcΔT, where q is the heat exchanged, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, we calculate the heat gained by the water:

q = m∙c∙ΔT
∙qwater = (50.0 mL ∙ 1g/mL) ∙ (4.18 J/g°C) ∙ (17.4°C – 15.0°C)
∙qwater = 502.0 g ∙ (4.18 J/g°C) ∙ (2.4°C)
∙qwater = 502.0 g ∙ (4.18 J/g°C) ∙ 2.4°C
∙qwater = 5024.32 J

Next, we use this value to find the mass of the iron pin:

∙qiron = (-)∙qwater
∙miron ∙ ciron ∙ (ΔTiron) = -5024.32 J
∙miron = 5024.32 J / (0.473 J/g°C ∙ 92.6°C)
∙miron = 5024.32 J / (43.788 J/g)
∙miron = 114.73 g

The mass of the iron pin is approximately 114.73 grams.