Question

Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water at 35.0°C. The final temperature of the mixture is 45.0°C. What is the specific heat of the metal in the weights?

Answer 1

Answer:

Approximately $0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity $c$ and mass $m$. Increasing the temperature of this object by $\Delta T$ would require $Q = c\, m \, \Delta T$.

Look up the specific heat of water: $c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}$.

It is given that the mass of the water in this mixture is $m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}$.

Temperature change of the water: $\Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}$.

Thus, the water in this mixture would have absorbed :

$\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}$.

Thus, the energy that water absorbed was: $Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}$.

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, $Q(\text{water})$, would be the opposite of the energy that the metal in this mixture released.

Thus: $Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J}$ (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: $m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}$.

Temperature change of the metal in this mixture: $\Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}$.

Rearrange the equation $Q = c\, m \, \Delta T$ to obtain an expression for the specific heat capacity: $c = Q / (m\, \Delta T)$. The (average) specific heat capacity of the metal pieces in this mixture would be:

$\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}$.