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

To calculate the final temperature in this scenario, we need to use the principle of heat transfer and the equation for heat gained or lost. By calculating the heat gained by the water and the aluminum cup, we can determine the final temperature. Substitute the known values into the equation and solve for the final temperature.

Step-by-step explanation:

To calculate the final temperature when a copper block is placed in an aluminum calorimeter cup containing water, we need to use the principle of heat transfer.

The heat lost by the copper block is equal to the heat gained by the water and the aluminum cup. The equation for heat transfer is:

Heat lost by copper = Heat gained by water + Heat gained by aluminum cup

We can use the equation Q = mcΔT to calculate the heat gained or lost. By rearranging the equation, we can solve for the final temperature.

First, calculate the heat gained by the water: Q_water = m_water * c_water * ΔT_water, where m_water is the mass of water, c_water is the specific heat capacity of water, and ΔT_water is the change in temperature of the water.

Next, calculate the heat gained by the aluminum cup: Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum, where m_aluminum is the mass of the aluminum cup, c_aluminum is the specific heat capacity of aluminum, and ΔT_aluminum is the change in temperature of the aluminum cup.

Using the principle of heat transfer and the equation for heat gained or lost, we can solve for the final temperature. Substitute the known values into the equation and solve for the final temperature.

Answer 2

T = 20.84°C

Step-by-step 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 = (78560935\ J)/(3770145\ J/^oC)`

T = 20.84°C