# 30 grams of a metal at 100 degrees C is placed in a calorimetre containing 100 mL of water at 21...

## Question:

{eq}30 \ grams {/eq} of a metal at {eq}100^\circ \ C {/eq} is placed in a calorimetre containing {eq}100 \ mL {/eq} of water at {eq}21^\circ \ C. {/eq} After a few minutes, the temperature inside the calorimetre is {eq}24^\circ \ C {/eq}. What is the specific heat of the metal?

## Heat Transfer Equation:

The heat transfer equation holds true when no physical transformation or chemical reaction occurs during the process. The heat transfer equation is given as {eq}\displaystyle q = mc\Delta T {/eq}. In this equation, m is the mass of the substance, c is the specific heat, and {eq}\displaystyle \Delta {/eq}T is the resulting change in the temperature of the substance.

## Answer and Explanation:

Determine the specific heat of the metal by equating the heat lost by the metal, {eq}\displaystyle -q_{metal} {/eq}, to the heat gained by the water, {eq}\displaystyle q_{water} {/eq}. Thus, we follow the general equation, {eq}\displaystyle -q_{metal} = q_{water} {/eq}. We use the heat transfer equation, {eq}\displaystyle q = mc\Delta T {/eq}, to help us solve the problem. We assume that water has a density of 1 g/mL We use the following values:

• {eq}\displaystyle m_{metal} = 30\ g {/eq}
• {eq}\displaystyle \Delta T_{metal} = 24 - 100 = -76 ^\circ C {/eq}
• {eq}\displaystyle m_{water} = 100 \ mL \times 1\ g/mL = 100\ g {/eq}
• {eq}\displaystyle c_{water} = 4.18 \ J/ g ^\circ C {/eq}
• {eq}\displaystyle \Delta T_{water} = 24 - 21 = 3 ^\circ C {/eq}

We proceed with the solution.

{eq}\begin{align} \displaystyle -q_{metal} &= q_{water}\\ -m_{metal}c_{metal}\Delta T_{metal} &= m_{water}c_{water}\Delta T_{water}\\ c_{metal} &= -\frac{m_{water}c_{water}\Delta T_{water}}{m_{metal}\Delta T_{metal}}\\ &= -\frac{100\ g\times 4.18 \ J/ g ^\circ C\times 3 ^\circ C}{30\ g\times -76 ^\circ C}\\ &= \boxed{\rm 0.6\ J/g ^\circ C} \end{align} {/eq}