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

1.90 x 10² cal

1.90 x 10² cal

0.133 cal/(g °C)

Explanation:

For water :

= mass of water = 112 g

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

= initial temperature of water = 90.0 °C

= final temperature of water = 88.3 °C

= Heat lost by water

Heat lost by water is given as

= (112) (1) (90.0 - 88.3)

= 1.90 x 10² cal

= Heat gained by the block

As per conservation of energy

Heat gained by the block = Heat lost by water

=

= 1.90 x 10² cal

For Block :

= mass of block = 21.0 g

= specific heat of block

= initial temperature of block = 20.0 °C

= final temperature of block = 88.3 °C

= Heat gained by Block = 1.90 x 10² cal

Heat gained by water is given as

= 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)