A 237 g piece of molybdenum, initially at 100.0°C, is dropped into 244 g of water at 10.0°C. When the system comes to thermal equilibrium, the temperature is 15.3°C. What is the specific heat capacity of molybdenum?

0.2695J

heat lost by Mo + heat gained by H2O = 0

[mass Mo x specific heat Mo x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)]= 0
specific heat Mo is the only unknown.

To find the specific heat capacity of molybdenum, we can use the principle of conservation of energy. The heat gained by the water will be equal to the heat lost by the molybdenum.

First, let's calculate the heat gained by the water:
Q_water = m_water * c_water * ΔT_water
where:
Q_water is the heat gained by the water,
m_water is the mass of water,
c_water is the specific heat capacity of water,
ΔT_water is the change in temperature of the water.

Given:
m_water = 244g
c_water (specific heat capacity of water) = 4.18 J/g°C
ΔT_water = 15.3°C - 10.0°C = 5.3°C

Q_water = 244g * 4.18 J/g°C * 5.3°C
Q_water = 5433.832 J

Now, let's calculate the heat lost by the molybdenum:
Q_molybdenum = m_molybdenum * c_molybdenum * ΔT_molybdenum
where:
Q_molybdenum is the heat lost by the molybdenum,
m_molybdenum is the mass of the molybdenum,
c_molybdenum is the specific heat capacity of molybdenum,
ΔT_molybdenum is the change in temperature of the molybdenum.

Given:
m_molybdenum = 237g
ΔT_molybdenum = 15.3°C - 100.0°C = -84.7°C (negative because the temperature decreased)

Q_molybdenum = 237g * c_molybdenum * -84.7°C

Since the two substances came to thermal equilibrium, the heat gained by the water is equal to the heat lost by the molybdenum:
Q_water = Q_molybdenum

5433.832 J = 237g * c_molybdenum * -84.7°C

Now, we can solve for c_molybdenum by rearranging the equation:

c_molybdenum = 5433.832 J / (237g * -84.7°C)

Calculating this, we get:

c_molybdenum ≈ 0.272 J/g°C

So, the specific heat capacity of molybdenum is approximately 0.272 J/g°C.

e213