Respuesta :
Answer:
The specific heat capacity of the metal is 1.307 J/g °C
Explanation:
Step 1: Data given
mass of the metal = 312 grams
initial temperature of the metal ( before plunged in the water) = 277.845 °C
initial temperature water = 11.945 °C
Final temperature of water (and aslo the metal) = 99.062 °C
Specific heat capacity of wayer = 4.184 J/g °C
Step 2: Calculate the specifi heat capacity of the metal
Loss of Heat of the Metal = Gain of Heat by the Water
Qmetal = -Qwater
m(metal) * C(metal) * ΔT(metal) = - m(water) * C(water) * ΔT(water)
with mass of metal = 312 grams
with C(metal) = TO BE DETERMINED
with ΔT (metal) = 99.062 - 277.845 = -178.783 °C
with mass of water = 200 grams
with C(water) = 4.184 J/°C * g
with ΔT (water) = 99.062 - 11.945 = 87.117 °C
312 * C(metal) * -178.783 = - 200* 4.184 * 87.117
C(metal) = -72899.51 / 55780.296 = 1.307
The specific heat capacity of the metal is 1.307 J/g °C
The specific heat capacity of the metal is 1.307 J/g °C.
Specific heat capacity
It is the ratio of the amount of heat required to raise the temperature of a substance by one degree Celsius to the amount of heat required to raise the temperature of the equal amount of water by one-degree Celsius at the room temperature.
[tex]Q=m*c*\triangle T[/tex]
Given:
Mass of the metal, m= 312 grams
Initial temperature of the metal ( before plunged in the water) = 277.845 °C
initial temperature water = 11.945 °C
Final temperature of water (with metal) = 99.062 °C
Specific heat capacity of water ,c = 4.184 J/g °C
→ Calculation for the specific heat capacity of the metal:
Loss of Heat of the Metal = Gain of Heat by the Water
[tex]Q_{metal} = -Q_{water}[/tex]
m(metal) * C(metal) * ΔT(metal) = - m(water) * C(water) * ΔT(water)
[tex]312 * C_{(metal)} * -178.783 = - 200* 4.184 * 87.117\\\\C_{(metal)} = \frac{-72899.51}{55780.296 }= 1.307[/tex]
The specific heat capacity of the metal is 1.307 J/g °C.
Find more information about Specific heat here:
brainly.com/question/13439286
