Question

PLEASE HELP!

A piece of metal is heated to a temperature of 50.0°C and then placed in a calorimeter containing 50.0 g of water at 27.0°C. The water temperature increases to 35.0°C. How many joules of heat were transferred from the metal to the water? (Cwater = 4.18J/g•°C)

Answer 1

Answer: 1672 Joules

Step-by-step explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

`Q=m* c* \Delta T`

Q = Heat absorbed by water = ?

c = heat capacity of water =
`4.18 J/g^0C`

Initial temperature of the water =
`T_i` =
`27^0C`

Final temperature of the water =
`T_f` =
`35^0C`

Change in temperature ,
`\Delta T=T_f-T_i=(35.0-27.0)^0C=8.0^0C`

Putting in the values, we get:

`Q=50.0g* 4.18J/g^0C* 8.0^0C=1672J`

As heat absorbed by water is equal to the heat released by metal, the Joules of heat transferred are 1672 Joules