A block of titanium (‘c’ = 0.54 J/ g o C ) weighing 6 5 g and at 1 0 5 o C was d
ID: 589728 • Letter: A
Question
A block of titanium (‘c’ = 0.54 J/goC) weighing 65 g and at 105 oC was dropped into 75.0 g of water kept at a room temperature of 22.6 oC in a calorimeter. Calculate the final temperature that will be reached by the metal and water? A block of titanium (‘c’ = 0.54 J/goC) weighing 65 g and at 105 oC was dropped into 75.0 g of water kept at a room temperature of 22.6 oC in a calorimeter. Calculate the final temperature that will be reached by the metal and water? A block of titanium (‘c’ = 0.54 J/goC) weighing 65 g and at 105 oC was dropped into 75.0 g of water kept at a room temperature of 22.6 oC in a calorimeter. Calculate the final temperature that will be reached by the metal and water? A block of titanium (‘c’ = 0.54 J/goC) weighing 65 g and at 105 oC was dropped into 75.0 g of water kept at a room temperature of 22.6 oC in a calorimeter. Calculate the final temperature that will be reached by the metal and water? A block of titanium (‘c’ = 0.54 J/goC) weighing 65 g and at 105 oC was dropped into 75.0 g of water kept at a room temperature of 22.6 oC in a calorimeter. Calculate the final temperature that will be reached by the metal and water?Explanation / Answer
Heat lost by Titanium =heat gained by water
Let the final temperature is Tf
Heat = m*C*dt
m = mass
C= heat capcity
dT = temperature change
65*0.54*(105- Tf) = 75*4.18*(Tf - 22.6)
- 35.1 Tf + 3685.5 = 313.5 Tf - 7085.1
Tf = 30.896 C
Thr final temeperature is 30.896 C
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