Question

A 0.0675 kg ingot of metal is heated to 242 ◦C and then is dropped into a beaker containing 0.374 kg of water with an initial temperature of 23 ◦C. If the final equilibrium state of the mixed system is 25.4 ◦C, find the specific heat of the metal. The specific heat of water is 4186 J/kg ·◦ C. Answer in units of J/kg ·◦ C.

Answer 1

To calculate the specific heat of the metal, we can use the principle of conservation of energy, which states that the heat lost by the metal will be equal to the heat gained by the water.

Let:
- \( m_m = 0.0675 \) kg (mass of the metal)
- \( c_m \) = specific heat of the metal (in J/kg·°C, the value we want to find)
- \( T_{m_i} = 242 \) °C (initial temperature of the metal)
- \( T_f = 25.4 \) °C (final equilibrium temperature)

- \( m_w = 0.374 \) kg (mass of the water)
- \( c_w = 4186 \) J/kg·°C (specific heat of water)
- \( T_{w_i} = 23 \) °C (initial temperature of the water)

Using the heat loss of the metal and the heat gain of the water:

\[
Q_{lost \, by \, metal} = Q_{gained \, by \, water}
\]

The heat lost by the metal is given by:
\[
Q_{metal} = m_m \cdot c_m \cdot (T_{m_i} - T_f)
\]
The temperature change for the metal is \( T_{m_i} - T_f = 242 - 25.4 \).

The heat gained by the water is given by:
\[
Q_{water} = m_w \cdot c_w \cdot (T_f - T_{w_i})
\]
The temperature change for the water is \( T_f - T_{w_i} = 25.4 - 23 \).

Setting the two heat equations equal to each other:

\[
m_m \cdot c_m \cdot (T_{m_i} - T_f) = m_w \cdot c_w \cdot (T_f - T_{w_i})
\]

Plugging in the values:

\[
0.0675 \cdot c_m \cdot (242 - 25.4) = 0.374 \cdot 4186 \cdot (25.4 - 23)
\]

Now, calculate the temperature changes:

\[
T_{m_i} - T_f = 242 - 25.4 = 216.6 \, °C
\]
\[
T_f - T_{w_i} = 25.4 - 23 = 2.4 \, °C
\]

Now substituting these values into the equation:

\[
0.0675 \cdot c_m \cdot 216.6 = 0.374 \cdot 4186 \cdot 2.4
\]

Calculating the right side:

\[
0.374 \cdot 4186 \cdot 2.4 = 0.374 \cdot 10046.4 = 3755.5936 \, J
\]

Now rewritting our equation:

\[
0.0675 \cdot c_m \cdot 216.6 = 3755.5936
\]

Now, divide both sides by \( 0.0675 \cdot 216.6 \):

\[
c_m = \frac{3755.5936}{0.0675 \cdot 216.6}
\]

Calculating \( 0.0675 \cdot 216.6 \):

\[
0.0675 \cdot 216.6 = 14.6211
\]

Now substituting back in:

\[
c_m = \frac{3755.5936}{14.6211} \approx 256.58 \, J/kg \cdot °C
\]

Thus, the specific heat of the metal is approximately:

\[
\boxed{256.58} \, \text{J/kg·°C}
\]