Question

A volume of 38.7 mL of H2O is initially at 28.0 oC. A chilled glass marble weighing 4.00 g with a heat capacity of 3.52 J/oC is placed in the water. If the final temperature of the system is 26.1 oC , what was the initial temperature of the marble? Water has a density of 1.00 g/mL and a specific heat of 4.18 J/goC. Enter your answer numerically, to three significant figures and in terms of oC.

Answer 1

- 61.2°C = Initial T°

Step-by-step explanation:

This is a calorimetry problem, where the heat from the water was gained by the marble.

Q = m . C . ΔT

where ΔT is final T° - initial T°, C is the specific heat and m, mass.

By the volume of water, we realize the mass (we apply density):

1 g/mL = mass / 38.7 mL

38.7 g = mass of water

Now, we need to find out the specific heat for the marble and we have Heat Capacity data

Heat Capacity = C . m

3.52 J/°C / 4 g = C → specific heat → 0.88 J/g °C

We make the equations for both heats:

m . C . ΔT from water = m . C . ΔT from the marble

38.7 g . 4.18 J/g°C ( 26.1°C - 28°C) = 4 g . 0.88 J/g °C . (26.1°C - Initial T°)

307.35 J = 4 g . 0.88 J/g °C . (26.1°C - Initial T°)

- 307.35 is a negative value, because the was has decreased the temperature, is a loss of heat, but we have to work with the positive number.

307.35 J / (4 . 0.88 °C/J) = 26.1°C - Initial T°

87.32°C = 26.1°C - Initial T°

- 61.2°C = Initial T°