Physics, 23.07.2019 13:00 craycrayjayjay123

Three lead fishing weights, each with a mass of 1.00x10^2 g and at a temperature of 100.0oc, are placed in 2.00x10^2 g of water at 35.0oc. the final temperature of the mixture is 45.0oc. what is the specific heat of the lead in the weights?

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The lead and the water reach thermal equilibrium when the net flow of heat from the lead to the water becomes zero, so when the heat released by the lead Q_L

is equal to the heat absorbed by the water Q_w

(1)
where the negative sign means the heat of the lead is released, not absorbed.

The equation that gives the heat absorbed/released is
$Q=m C \Delta T$ (2)
where
m is the mass of the substance
C is its specific heat capacity
$\Delta T$ is the variation of temperature.

If we rewrite (1) by using (2), we get
$-m_L C_L \Delta T_L = m_w C_w \Delta T_w$ (3)
where the label L refers to the lead and the label W refers to the water.

The total mass of the lead is:
$m_L = 3 \cdot 1.00 \cdot 10^2 g =300 g$
while its change of temperature is
$\Delta T_L = T_f - T_i = 45.0^{\circ} C - 100^{\circ}C = -55^{\circ} C$

The mass of the water is
$m=2.00 \cdot 10^2 g = 200 g$
its specific heat capacity is
$C_w = 4.18 J/(g ^{\circ}C)$
while its change of temperature is
$\Delta T_w = T_f - T_i = 45.0^{\circ} C- 35.0^{\circ} C= 10^{\circ}C$

If we put all this data inside (3), we can calculate the specific heat capacity of lead:
$C_L = - \frac{m_w C_w \Delta T_w}{m_L C_L}=- \frac{(200 g)(4.18 J/g^{\circ}C)(10^{\circ}C)}{(300 g)(-55^{\circ}C)}= 0.50 J/(g^{\circ}C)$

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Three lead fishing weights, each with a mass of 1.00x10^2 g and at a temperature of 100.0oc, are pla...

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