Question

A chunk of aluminum at 91.4°C was added to 200.0 g of water at 15.5°C. The specific heat of aluminum is 0.897 J/g°C, and the specific heat of water is 4.18 J/g°C. When the temperature stabilized, the temperature of the mixture was 18.9°C. Assuming no heat was lost to the surroundings, what was the mass of aluminum added?

Answer 1

Answer:

$m_a=931.9955\ g$

Explanation:

Given:

• initial temperature of Aluminium, $T_a_i=91.4^{\circ}C$

• mass of water, $m=200\ g$

• initial temperature of water, $T_w_i=15.5^{\circ}C$

• specific heat of aluminium, $c_a=0.897\ J.g^{-1}.^{\circ}C^{-1}$

• specific heat of water, $c_w=4.18\ J.g^{-1}.^{\circ}C^{-1}$

• final temperature of the mixture, $T_f=18.9^{\circ}C$

Heat lost by Al to go from 91.4°C to 18.9°C:

$Q_a=m_a.c_a.\Delta T_a$

where:

$\Delta T_a=$ temperature difference for Al

$m_a=$ mass of aluminium

$\Rightarrow Q_a=m_a\times 0.897\times 72.5$

$Q_a=65.0325\times m_a$ ...........................................(1)

Heat gained by water to go from 15.5°C to 18.9°C:

$Q_w=m_w.c_w.\Delta T_w$

where:

$\Delta T_w=$ temperature difference for water

$\Rightarrow Q_w=200\times 4.18\times 72.5$

$Q_w=60610\ J$ ...........................................(2)

According to the law of conservation of energy:

When there is no heat loss to the surrounding:

$Q_a=Q_w$

$65.0325\times m_a=60610$

$m_a=931.9955\ g$