Question

A piece of metal weighing 59.047 g was heated to 100.0 °C and then put it into 100.0 mL of water (initially at 23.7 °C). The metal and water had final temp of 27.8 °C. What is the specific heat of the metal?

Answer 1

Answer: The specific heat of metal is 0.403J/g°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 100.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{100.0mL}\\\\\text{Mass of water}=(1g/mL\times 100.0mL)=100g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$       ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 59.047 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 27.8°C

$T_1$ = initial temperature of lead = 100°C

$T_2$ = initial temperature of water = 23.7°C

$c_1$ = specific heat of lead = ?

$c_2$ = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$59.047\times c_1\times (27.8-100)=-[100\times 4.186\times (27.8-23.7)]$

$c_1=0.403J/g^oC$

Hence, the specific heat of metal is 0.403J/g°C

Answer 2

The specific heat of the metal is equal to 0.4024 J/g°C.

Given the following data:

• Mass of metal = 59.047 g

• Initial temperature of metal = 100.0°C

• Initial temperature of water = 23.7°C

• Final temperature of water = 27.8°C

• Final temperature of metal = 27.8°C

• Volume of water = 100.0 mL

Specific heat of water = 4.184 J/g°C

Density of water = 1 g/mL

To determine the specific heat of the metal:

First of all, we would determine the mass of water.

$Mass = density \times volume\\\\Mass = 1 \times 100$

Mass = 100 grams

The quantity of heat lost by the piece of metal = The quantity of heat gained by the water.

$Q_{lost} = Q_{gained}\\\\mc\theta = mc\theta\\\\59.047c(100-27.8) = 100(4.184)(27.8 -23.7)\\\\59.047c(72.2) = 418.4(4.1)\\\\4263.1934c = 1715.44\\\\c = \frac{1715.44}{4263.1934}$

Specific heat, c = 0.4024 J/g°C

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