A 50 g sample of an unknown metal is heated to 90.0C. It is placed in a perfectly insulated container along with 100 g of water at an initial temperature of 20C. After a short time, the temperature of both the metal and water become equal at 25C. The specific heat of water is 4.18 J/gC in this temperature range. What is the specific heat capacity of the metal?

Record your answer with two significant figures.

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To find the specific heat capacity of the metal, we can use the principle of heat transfer. The heat gained or lost by an object can be calculated using the formula:

Q = m * c * ΔT

where:
Q is the heat gained or lost
m is the mass of the object
c is the specific heat capacity of the object
ΔT is the change in temperature

In this case, we observe that the metal and water reach a final temperature of 25°C, which means they gained or lost the same amount of heat.

First, we can calculate the heat gained by the water using the formula above:

Q_water = m_water * c_water * ΔT_water

where:
m_water is the mass of the water (100 g)
c_water is the specific heat capacity of water (4.18 J/g°C)
ΔT_water is the change in temperature (-5°C, since it went from 20°C to 25°C)

Q_water = (100 g) * (4.18 J/g°C) * (-5°C)
Q_water = -2090 J

Since the metal and water share the same final temperature, the heat gained by the metal is equal in magnitude but opposite in sign:

Q_metal = -Q_water
Q_metal = 2090 J

Now, we can calculate the specific heat capacity of the metal using the heat gained by the metal:

Q_metal = m_metal * c_metal * ΔT_metal

where:
m_metal is the mass of the metal (50 g)
c_metal is the specific heat capacity of the metal (what we want to find)
ΔT_metal is the change in temperature (25°C - 90°C = -65°C)

Substituting the known values, we can solve for c_metal:

2090 J = (50 g) * c_metal * (-65°C)
2090 J = -3250 g°C * c_metal
c_metal = 2090 J / (-3250 g°C)
c_metal ≈ -0.642 J/g°C

Rounding to two significant figures, the specific heat capacity of the metal is approximately -0.64 J/g°C.