Question

A 175 gram sample of a metal at 93.50c was added to 105 grams of water at 23.50c in a perfectly insulated container. the final temperature of the water and metal was 33.80c. calculate the specific heat of the metal in j/g0c.

Answer 1

To calculate the specific heat of the metal, use the formula q = mc∆T. Set up the equation mc(Metal)∆T(Metal) = mc(Water)∆T(Water). Solve for the specific heat of the metal using the given values.

Step-by-step explanation:

To calculate the specific heat of the metal, we can use the formula: q = mc∆T, where q is the heat gained or lost, m is the mass, c is the specific heat, and ∆T is the change in temperature. In this case, the heat gained by the metal is equal to the heat lost by the water. We can set up the equation: mc(Metal)∆T(Metal) = mc(Water)∆T(Water). Since the metal and water came to the same temperature, ∆T(Metal) = ∆T(Water). Solving for the specific heat of the metal, we get:

c(Metal) = (m(Water) x c(Water) x ∆T(Water)) / (m(Metal) x ∆T(Metal)).

Plugging in the given values:

c(Metal) = (105 g x 4.18 J/g°C x (33.80°C - 23.50°C)) / (175 g x (33.80°C - 93.50°C)).

Simplifying:

c(Metal) = 6.72 J/g°C.

Answer 2

The specific heat capacity (in J/gºC) of the 175 grams metal which was added to the water is 0.433 J/gºC

How to solve for the specific heat capacity of the metal?

From the above question, we were informed that the metal at 93.50 °C was added to the water at 23.50 °C. This implies that heat energy will be released by the metal into the water.

Now we shall calculate the heat energy released into the water. Details below:

• Mass of water (M) = 105 grams
• Initial temperature (T₁) = 23.50 °C
• Final temperature (T₂) = 22.31 °C
• Change in temperature (ΔT) = 33.80 - 23.50 = 10.30 °C
• Specific heat capacity of water (C) = 4.184 J/gºC
• Heat released into the water (Q) =?

Heat released = -heat absorbed

But

Heat absorbed = MCΔT

= 105 × 4.184 × 10.3

= 4524.996 J

Thus,

Heat released = -4524.996 J

Now, we shall calculate the specific heat capacity. This is shown below for better understanding:

• Heat released = -4524.996 J
• Mass of meta (M) = 175 grams
• Initial temperature of metal (T₁) = 93.50 °C
• Final temperature of metal (T₂) = 33.80 °C
• Change in temperature of metal (ΔT) = 33.80 - 93.50 = -59.7 °C
• Specific heat capacity (C) = ?

`C = (Heat\ released)/(M\Delta T) \\\\C = (-4524.996)/(175\ *\ -59.7) \\\\C = 0.433\ J/g\textdegree C`