Question

A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23°C. What is the specific heat of the metal?

Answer 1

Answer:

$0.8696\ \text{J/g}^{\circ}\text{C}$

Explanation:

$m_m$ = Mass of metal = 19 g

$c_m$ = Specific heat of the metal

$\Delta T_m$ = Temperature difference of the metal = $99-23=76^{\circ}\text{C}$

V = Volume of water = 150 mL = $150\ \text{cm}^3$

$\rho$ = Density of water = $1\ \text{g/cm}^3$

$c_w$ = Specific heat of the water = 4.186 J/g°C

$\Delta T_w$ = Temperature difference of the water = $23-21=2^{\circ}\text{C}$

Mass of water

$m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}$

Heat lost will be equal to the heat gained so we get

$m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}$

The specific heat of the metal is $0.8696\ \text{J/g}^{\circ}\text{C}$.