1. When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?
2. A calorimeter contains 100 g of water at 39.8 ºC. A 8.23 g object at 50 ºC is placed inside the calorimeter. When equilibrium has been reached, the new temperature of the water and metal object is 40 ºC. What type of metal is the object made from?
PLEASE SHOW WORK SO I CAN UNDERSTAND!!!!
Solution
1) Q= C x M x T
Q= energy in (joules)
C= heat capacity ( joules/gramsCelcius ) C of water(h20) is 4.18 j/gc
M = mass ( grams)
T= Temperature change ( in celcius)
3kg= 3000g
4.18j/gc x 3000g x 70c = 877800 Joules or 877.8 Kj or Calories
and it can be interpreted that you require 877.8 KiloJoules of energy to warm 3kg of water from 10C to 80C
2) Thermal energy is conserved. Heat lost by the hot object is gained by the cold object. Assumptions must be made. The only two \"objects\" are the \"8.23g object\" and water. The heat gained by the calorimeter itself is ignored. Heat is computed from: q=mc?T. (To make the calculations meaningful, three significant digits are assumed for all measurements.)
q(lost) = -q(gained)
mc?T(object) = -mc?T(water)
c(object) = -mc?T(water) / m?T(object)
c(object) = (-100.g x 4.18J/gC x 0.200C) / (8.23g x -10.0C)
c(object) = 1.015 J/gC
The only substance which has a specific heat capacity close to the calculated value is Magnesium (1.0467)
.
1- When 3-0 kg of water is warmed from 10 C to 80 C- how much heat ene.docx
1. 1. When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?
2. A calorimeter contains 100 g of water at 39.8 ºC. A 8.23 g object at 50 ºC is placed inside
the calorimeter. When equilibrium has been reached, the new temperature of the water and metal
object is 40 ºC. What type of metal is the object made from?
PLEASE SHOW WORK SO I CAN UNDERSTAND!!!!
Solution
1) Q= C x M x T
Q= energy in (joules)
C= heat capacity ( joules/gramsCelcius ) C of water(h20) is 4.18 j/gc
M = mass ( grams)
T= Temperature change ( in celcius)
3kg= 3000g
4.18j/gc x 3000g x 70c = 877800 Joules or 877.8 Kj or Calories
and it can be interpreted that you require 877.8 KiloJoules of energy to warm 3kg of water from
10C to 80C
2) Thermal energy is conserved. Heat lost by the hot object is gained by the cold object.
Assumptions must be made. The only two "objects" are the "8.23g object" and water. The heat
gained by the calorimeter itself is ignored. Heat is computed from: q=mc?T. (To make the
calculations meaningful, three significant digits are assumed for all measurements.)
q(lost) = -q(gained)
mc?T(object) = -mc?T(water)
c(object) = -mc?T(water) / m?T(object)
c(object) = (-100.g x 4.18J/gC x 0.200C) / (8.23g x -10.0C)
c(object) = 1.015 J/gC
2. The only substance which has a specific heat capacity close to the calculated value is Magnesium
(1.0467)
