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

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

Step-by-step 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 * 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 * 1.00 * 10^(2)\; {\rm g} * 10\; {\rm K} \\ &= 4.182 * 10^(3)\; {\rm J}\end{aligned}`.

Thus, the energy that water absorbed was:
`Q(\text{water}) = 4.182 * 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 * 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 * 1.00 * 10^(2)\; {\rm g} = 3.00 * 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 &= (Q)/(m\, \Delta T) \\ &= \frac{-4.182 * 10^(3)\; {\rm J}}{3.00 * 10^(2)\; {\rm g} * (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^(-1) \cdot K^(-1)}\end{aligned}`.