Part A The value of specific heat for copper is 390 J/kgC, for aluminun is 900 J
ID: 1509296 • Letter: P
Question
Part A
The value of specific heat for copper is 390 J/kgC, for aluminun is 900 J/kgC, and for water is 4186 J/kgC.
What will be the equilibrium temperature when a 275 g block of copper at 255 C is placed in a 155 galuminum calorimeter cup containing 815 g of water at 16.0 C?
Part B
A 0.40-kg iron horseshoe, just forged and very hot (Figure 1) , is dropped into 1.45 L of water in a 0.33-kgiron pot initially at 20.0C. The value of specific heat for iron is 450 J/kgC , and for water is 4186 J/kgC.
If the final equilibrium temperature is 25.0 C, Determine the initial temperature of the hot horseshoe.
Part C
High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kgC, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kgC.
Calculate the energy absorbed from a climber's body if he eats 0.70 kg of -15C snow which his body warms to body temperature of 37C.
Part D
Calculate the energy absorbed from a climber's body if he melts 0.70 kg of -15C snow using a stove and drink the resulting 0.70 kg of water at 2C, which his body has to warm to 37C.
Explanation / Answer
These exercises can be performed using the expression of calorimetry
Qcool = Q heat
Q = m ce (Tf-Ti)
Part A)
Qheat = mcu Cecu (( Ticu -Teq)
Qcool = mal Ceal (Tial – Teq) + mwater Cewater ( Ted - Ti water)
mal Ceal (Tial – Teq) + mwater Cewater ( Ted - Ti water) = -mcu Cecu (( Ticu -Teq)
0.155 900 ( Teq – 16 ) + 0.815 4186 (Teq – 16) = - 0.275 390 ( Teq – 255)
(139.5 + 3411.59 ) (Teq -16) = 107.25 (255 – Teq)
3551 (Teq – 16) = 107.25 255 - 107.25 Teq
3551 Teq - 16 3551 = 107.25 255 - 107.25 Teq
3551 Teq + 107.25 Teq = 27348.75 + 56816
Teq (3551 + 107.25) = 84164.75
Teq = 84164.75/3658.25
Teq = 23 C
Part B)
0.40 450 ( Ti – 25.0) = 1.45 4186 (25-20) + 0.33 450 (25-20)
180 (Ti -25) = 6069.7 5+148.5 5
180 Ti – 180 25 = 30348.5 + 742.5
180 Ti = 31091 + 4500
Ti = 35591/180
Ti =197.73 C
Part C)
The absorbed heat is the heat calenter ice to zero degree Celsius, but the heat of the ice deretir more heat to heat the water to body temperature
Qt = Q1 + Q2 + Q3
Q1 = m Ceice (Ti-Tf)
Q1 = 0.70 2100 ( 0 -(-15))
Q1 = 22050 J
Q2 = m L
Q2 = 0.7 333
Q2 = 233.1 J
Q3 = m Cewater (Tf-Ti)
Q3 = 0.7 4186 (37 – 0)
Q3 = 108417.4 J
Qt = 22050 + 233.1 + 108417.4
Qt= 130700.5 J
Part D)
the absorbed heat is to heat the water to 37C
Q = 0.7 4186 (37 -2)
Q= 102557 J