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Alinara [238K]
3 years ago
15

A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

24.0oC. Neglecting the heat capacity of the container, find the final equilibrium temperature (in oC) of the lead and water.
Physics
1 answer:
Vaselesa [24]3 years ago
3 0

Answer:24.70 ^{\circ}C

Explanation:

Given

mass of lead piece m_l=234 gm\approx 0.234 kg

mass of water in calorimeter m_w=611 gm\approx 0.611 kg

Initial temperature of water T_w=24^{\circ}C

Initial temperature of lead piece T_l=24^{\circ}C

we know heat capacity of lead and water are 125.604 J/kg-k and 4.184 kJ/kg-k respectively

Let us take T ^{\circ}C be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

m_lc_l(T_l-T)=m_wc_w(T-T_w)

0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)

86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)

86-T=86.97T-2087.49

T=\frac{2173.491}{87.97}=24.70^{\circ}C

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