Question

A 0.350 kg block at -27.5°C is added to 0.217 kg of water at 25.0°C. They come to equilibrium at 16.4°C. What is the specific heat of the block? (Unit=J/(kg*C))

Answer 1

The specific heat capacity of the block is $508J/kg^{\circ}C$

Explanation:

As the block is placed into the water, heat energy is transferred from the water (which is at higher temperature) to the block (which is at lower temperature), until the block and the water are in thermal equilibrium (= same temperature).

Therefore, we can write:

$Q_{water}=Q_{block}$

Where

$Q_{water}=m_w C_w (T_w-T_{eq})$ is the heat energy released by the water, where

$m_w = 0.217 kg$ is the mass of the water

$C_w = 4186 J/kg^{\circ}C$ is the water heat specific capacity

$T_w = 25.0^{\circ}$ is the initial temperature of the water

$T_{eq}=16.4^{\circ}$ is the temperature at equilibrium

Substituting,

$Q_{water}=(0.217)(4186)(25.0-16.4)=7812 J$

Now we can write the heat energy absorbed by the block as

$Q_{block}=m_b C_b(T_{eq}-T_b)$

where

$m_b=0.350 kg$ is the mass of the block

$C_b$ is the specific heat capacity of the block

$T_b = -27.5^{\circ}$ is the initial temperature of the block

And solving for $C_b$,

$C_b=\frac{Q_{block}}{m_b(T_{eq}-T_b)}=\frac{7812}{(0.350)(16.4-(-27.5))}=508J/kg^{\circ}C$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly