Question

How do I solve this? Iron: 0.412 J/g/deg CWater: 4.1813 J/g/deg C

Answer 1

Given:

The mass of the iron is

`m_i=\text{ 352 g}`

The specific heat of iron is

`c_i=\text{ 0.412 J/g }^(\circ)C`

The initial temperature of the iron is

`T_i=\text{ 80.2 }^(\circ)C`

The mass of the water is

`m_w=\text{ 443 g}`

The specific heat of water is

`c_w=\text{ 4.1813 J/g }^(\circ)C`

The initial temperature of the water is

`T_w=\text{ 0}^(\circ)C`

Required: Final equilibrium temperature.

Step-by-step explanation:

Let the final equilibrium temperature be T.

The final equilibrium temperature can be calculated as

`\begin{gathered} m_ic_i(T-T_i)=m_wc_w(T-T_w) \\ 352*0.412(T-80.2)=443*4.1813(T-0) \\ 145.024T-11630.92=1852.32T \\ T=(-11630.92)/(1707.29) \\ =\text{ -6.81 }^(\circ)C \end{gathered}`