Question

1) You have a device like that shown in figure 14.3 of Joule's experiment (only that you only have one
weight). The mass is 1.0 kg, and it can fall 1.00 m. How much heat is added to 1.0 L of water?

2) How much does the temperature change in that 1.0 L of water?

3) Answer questions 1 and 2 if instead (of 1.0 L of water) you had a 100 g of liquid mercury?

4) How much heat is needed in order to completely boil away 2.5 L of water?

5) You have two ice cubes at 0 degrees Celsius, each 2.5 cm x 2.5 cm x 2.5 cm, that you put in 1.0 L of 
water at 25 degrees Celsius. What is the final temperature of the mix?

Answer 1

1] Heat added = mgh = 1*9.8*1 = 9.8 J

2] temperature change = heat given/[m*S] where m is mass of liquid and S is specific heat,

= 9.8/[1*4186]

= 0.00234 degree C

3] heat added will be same = 9.8 J

Temperature change = heat given/[m*S] where m is mass of liquid and S is specific heat,

= 9.8/[1*140]

= 0.07 degree C

4] To boil away water , heat required = m*S*deltaT + mL where m is mass of water, initial temperature is assumed to be 25C, L is latent heat of valorization,

heat = 2.5*4186*(100-25) + 2.5*2260e3

= 6434875 J = 6.43*10^6 J