Question

Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter cup containing 844 g of water at 14.6°C? ​

Answer 1

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

$m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a$

where,

$m_c$ = mass of copper = 227 g

$m_w$ = mass of water = 844 g

$m_a$ = mass of aluminum = 155 g

$C_c$ = specific heat capacity of calorimeter = 385 J/kg.°C

$C_w$ = specific heat capacity of water = 4200 J/kg.°C

$C_a$ = specific heat capacity of aluminum = 890 J/kg.°C

$\Delta T_c$ = change in temperature of copper = 283°C - T

$\Delta T_w$ = change in temperature of water = T - 14.6°C

$\Delta T_a$ = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

$(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}$

T = 20.84°C