A 125-g piece of metal is heated to 288 degrees celsius and dropped into 85.0 g of water at 26 degrees celsius. If the final temperature of water and metal is 58 degrees celsius, what is the specific heat of the metal?

0.129 J/g C

heat lost by metal + heat gained by water = 0

[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute the numbers and solve for the only unknown, the specific heat of the metal.

0.38J|g○C

700 9.76g

To find the specific heat of the metal, we need to use the principle of conservation of energy. The energy lost by the metal is equal to the energy gained by the water.

First, let's calculate the energy lost by the metal:
q1 = m1 * c1 * ΔT1
where:
- q1 is the heat lost or gained
- m1 is the mass of the metal
- c1 is the specific heat capacity of the metal
- ΔT1 is the change in temperature

For the metal, we have:
m1 = 125 g
c1 = ? (what we need to find)
ΔT1 = (final temperature - initial temperature) = (58°C - 288°C)

Next, let's calculate the energy gained by the water:
q2 = m2 * c2 * ΔT2
where:
- q2 is the heat lost or gained
- m2 is the mass of the water
- c2 is the specific heat capacity of water
- ΔT2 is the change in temperature

For the water, we have:
m2 = 85.0 g
c2 = 4.18 J/g°C (specific heat capacity of water)
ΔT2 = (final temperature - initial temperature) = (58°C - 26°C)

Since the energy lost by the metal is equal to the energy gained by the water (according to the principle of conservation of energy):
q1 = q2

Substituting the values we know into the equation, we have:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2

Now we can solve for c1:
c1 = (m2 * c2 * ΔT2) / (m1 * ΔT1)

Plugging in the values, we get:
c1 = (85.0 g * 4.18 J/g°C * (58°C - 26°C)) / (125 g * (58°C - 288°C))

Now it's just a matter of calculating the equation to find the specific heat of the metal.