A 50-gram sample of water is initially at a temperature of 22 °C. The sample is heated until the temperature is 32 °C The specific heat of water is 1.00 cal/g °C. How much heat is absorbed by the water in calories? Enter your answer without a decimal

500cal

Explanation:

Given parameters:

Mass of water  = 50g

Initial temperature  = 22°C

Final temperature  = 32°C

Specific heat of water  = 1cal/g

Unknown:

Amount of heat absorbed by the water in calories  = ?

Solution:

To solve this problem, we use the expression below:

       H  = m c Ф

H is the amount of heat absorbed

m is the mass

c is the specific heat capacity

Ф is the temperature change

       H  = 50 x 1 x (32  - 22)  = 500cal

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.