Question

1. an unknown piece of metal weighing 95 g is heated to 98°c. it is dropped into 250 g of water at 23°c. when equilibrium is reached, the temperature of the water and piece of metal is 29°c. determine the specific heat of the metal. the heat capacity of the container, a styrofoam cup, is negligible.

Explanation:

Step1: Determine the equilibrium temperature

The equilibrium temperature \(T_{eq}\) is \(29^{\circ}C\) (given as the final temperature of the water - metal mixture). The initial temperature of water \(T_{i,water}=23^{\circ}C\) and the initial temperature of metal \(T_{i,metal} = 98^{\circ}C\).

Step2: Use the heat - transfer formula

According to the principle of conservation of energy (assuming no heat loss to the surroundings, \(Q_{lost}=Q_{gained}\)). The heat transfer formula is \(Q = mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{final}-T_{initial}\).
For water, \(Q_{gained}=m_{water}c_{water}(T_{eq}-T_{i,water})\), where \(m_{water} = 250\space g\) and \(c_{water}=4.18\space J/g^{\circ}C\).
For metal, \(Q_{lost}=m_{metal}c_{metal}(T_{i,metal}-T_{eq})\), where \(m_{metal}=95\space g\)

Since \(Q_{lost} = Q_{gained}\), we have \(m_{metal}c_{metal}(T_{i,metal}-T_{eq})=m_{water}c_{water}(T_{eq}-T_{i,water})\)

Substitute the values:

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Answer:

The specific heat of the metal is \(0.96\space J/g^{\circ}C\)