Question

1. Calculate q when 29.6 g of ice is cooled from −26.2°C to −78.3°C (c_ice = 2.087 J/g·K)?

2. A 233−g aluminum engine part at an initial temperature of 11.77°C absorbs 80.4 kJ of heat. What is the final temperature of the part (c of Al = 0.900 J/g·K)?

3. A 24.0−g sample of ethylene glycol, a car radiator coolant, loses 510. J of heat. What was the initial temperature of the ethylene glycol if the final temperature is 32.5°C? (c of ethylene glycol = 2.42 J/g·K)?

Answer 1

1)

Solution-

Using the formula

q = m x Cp x ∆T

here m= mass

Cp= specific heat

∆T= change in temperature

So for change in temperature ∆T = (-26.2 + 273.15) - (-78.3 + 273.15)

= 52.1 K

= 29.6 g * (2.087 J / gK) * (52.1K))

= 3218.48 J

= -3218.48 J as the heat is being released and not absorbed