Thermal balance in calorimeter after adding lead

[tuki]

Homework Statement

In insulated copper container (mass 0.100 kg) there is water (mass 0.160 kg) and ice (mass 0.018 kg) in thermal balance. Lead is being added into container which is at 255 celcius and mass of 0.750kg. What is temperature when container and it's contents reach thermal balance after adding lead?

Some constants:
Copper thermal capacity: 390 J/kgK
Lead thermal capacity: 130 J/kgK
Water thermal capacity: 4190 J/kgK
Ice thermal capacity: 2100 J/kgK
Ice melting temperature: 334*10^3 J/kg
Lead melting point 327.3 Celicus

Relevant Equations

Thermal energy:
$$ Q = cm \Delta T $$

Since ΔT is change in temperature, the container and it's contents and the led most have same temperature difference when the led is added. I tried by assuming that energy released by the led is same as the amount that container and it's contents absorb. Meaning Q1-Q2 = 0 => Q1 = Q2.

$$ \Delta T(c_c m_c + c_w m_w) - m_i L = \Delta T(c_l m_l) $$
where _c = copper, _w = water, _i = ice, _l = led, L = Ice melting temperature.

By solving $\Delta T$ from this equation we should be able to figure how much temperature changes. However there are some things in this that don't add up. Now the temperature change should be equal for both except it doesn't? It cannot be equal since wouldn't make sense that both led and container would have same change in temperature.

Getting a little confused with this so maybe someone could help?

Correct answer should be approx 21 celicus

Also could someone tell me how to use latex / mathjax inside text. In mathexchange you can use two dollar signs $$ for math inside text and double for centered equation. However here the only two dollar signs won't work?

 

[kuruman]

You cannot use the same symbol $\Delta T$ for the two kinds of entities that change their temperature. The temperature change for the copper and ice is not the same as the temperature change for the lead. It's the final temperature of all three that is the same.

 

[tuki]

You cannot use the same symbol $\Delta T$ for the two kinds of entities that change their temperature. The temperature change for the copper and ice is not the same as the temperature change for the lead. It's the final temperature of all three that is the same.

Any ideas on how i could formulate this into equation? Could you give a hint?
I don't know how i can form equation with final temperature?

 

[kuruman]

Use $T_c~$, $T_w$ and $T_l$ for the known initial temperatures of copper, water and lead and $T_f$ for the common final temperature of all three. Then the temperature change for the lead would be $\Delta T_l=T_l-T_f$ and similar expressions for the other temperature changes in your equation. Substitute in your equation $T_l-T_f$ etc. and solve for $T_f$.

 

[DrClaude]

Also could someone tell me how to use latex / mathjax inside text. In mathexchange you can use two dollar signs $$ for math inside text and double for centered equation. However here the only two dollar signs won't work?

To insert a displayed equation, use $$ $$; to insert an inline equation, use ## ##.

 

[Chestermiller]

You had it almost right in your original post. The amount of heat received by the container, water, and ice is:
$$Q= m_iL+[c_c m_c + c_w( m_w+m_i)](T-0) $$This must match the amount of heat furnished by the lead:
$$Q=c_lm_l(255-T)$$