Question

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

Answer 1

1.90 x 10² cal

1.90 x 10² cal

0.133 cal/(g °C)

Step-by-step explanation:

For water :

`m_(w)` = mass of water = 112 g

`c_(w)` = specific heat of water = 1 cal/(g °C)

`T_(wi)` = initial temperature of water = 90.0 °C

`T_(wf)` = final temperature of water = 88.3 °C

`Q_(w)` = Heat lost by water

Heat lost by water is given as

`Q_(w)= m_(w)c_(w)(T_(wi) - T_(wf))`

`Q_(w)` = (112) (1) (90.0 - 88.3)

`Q_(w)` = 1.90 x 10² cal

`Q_(B)` = Heat gained by the block

As per conservation of energy

Heat gained by the block = Heat lost by water

`Q_(B)` =
`Q_(w)`

`Q_(B)` = 1.90 x 10² cal

For Block :

`m_(B)` = mass of block = 21.0 g

`c_(B)` = specific heat of block

`T_(bi)` = initial temperature of block = 20.0 °C

`T_(bf)` = final temperature of block = 88.3 °C

`Q_(B)` = Heat gained by Block = 1.90 x 10² cal

Heat gained by water is given as

`Q_(B)` = 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)