The temperature of a pot of water is 90.0 °C and the mass of the water is 112 g. A block with a mass of 21.0 g and a temperature of 20.0°C is dropped into the water. The final temperature of the water and the block is 88.3 °C. The specific heat of water is exactly 1 cal/g.°C. How much heat was lost by the water? O 1.90 x10^2 cal O 35.7 cal O 7650 cal O 1.70 cal How much heat was gained by the block? O 1430 cal O 190 x10^2 cal O 35.7 cal O 68.3 cal What is the specific heat of the block? O 0.133 cal/g.°C O 5.33 cal/g.°C O273000 cal/g.°C O0.0249 cal/g.°C

Question

contestada

Respuesta :

JemdetNasr JemdetNasr · Answer 1 · 2019-07-12T17:33:28+02:00

Answer:

1.90 x 10² cal

0.133 cal/(g °C)

Explanation:

For water :

[tex]m_{w}[/tex] = mass of water = 112 g

[tex]c_{w}[/tex] = specific heat of water = 1 cal/(g °C)

[tex]T_{wi}[/tex] = initial temperature of water = 90.0 °C

[tex]T_{wf}[/tex] = final temperature of water = 88.3 °C

[tex]Q_{w}[/tex] = Heat lost by water

Heat lost by water is given as

[tex]Q_{w}= m_{w}c_{w}(T_{wi} - T_{wf})[/tex]

[tex]Q_{w}[/tex] = (112) (1) (90.0 - 88.3)

[tex]Q_{w}[/tex] = 1.90 x 10² cal

[tex]Q_{B}[/tex] = Heat gained by the block

As per conservation of energy

Heat gained by the block = Heat lost by water

[tex]Q_{B}[/tex] = [tex]Q_{w}[/tex]

[tex]Q_{B}[/tex] = 1.90 x 10² cal

For Block :

[tex]m_{B}[/tex] = mass of block = 21.0 g

[tex]c_{B}[/tex] = specific heat of block

[tex]T_{bi}[/tex] = initial temperature of block = 20.0 °C

[tex]T_{bf}[/tex] = final temperature of block = 88.3 °C

[tex]Q_{B}[/tex] = Heat gained by Block = 1.90 x 10² cal

Heat gained by water is given as

[tex]Q_{B}[/tex] = m_{B}c_{B}(T_{bf} - T_{bi})[/tex]

1.90 x 10² = (21.0) (88.3 - 20.0) c_{B}

c_{B} = 0.133 cal/(g °C)