select the correct answer from each drop - do...
select the correct answer from each drop - down menu. a 126.1 - gram block of granite at 92.6°c is dropped into a tub of water at 24.7°c in an isolated system. the final temperature of both the granite and the water is 51.9°c. the specific heat capacity of granite is 0.795 joules/gram degree celsius, and the specific heat capacity of water is 4.186 joules/gram degree celsius. the granite block transferred of energy, and the mass of the water is
Answer
# Answer:
The granite block transferred 21,500 joules of energy.
Let the mass of water be $m$ grams.
Using the principle of conservation of energy (heat lost by granite = heat gained by water), we first find the heat lost by granite.
The formula for heat transfer is $Q = mc\Delta T$.
For granite:
$m_{granite}=126.1$ g, $c_{granite}=0.795$ J/g - °C, $\Delta T_{granite}=92.6 - 51.9=40.7$ °C
$Q_{lost - granite}=m_{granite}c_{granite}\Delta T_{granite}$
$Q_{lost - granite}=126.1\times0.795\times40.7$
$Q_{lost - granite}=126.1\times32.3565$
$Q_{lost - granite}\approx 4079.15\approx4080$ J
For water:
$c_{water} = 4.186$ J/g - °C, $\Delta T_{water}=51.9 - 24.7 = 27.2$ °C
Since $Q_{lost - granite}=Q_{gained - water}$
$Q_{gained - water}=m\times c_{water}\times\Delta T_{water}$
$4080=m\times4.186\times27.2$
$m=\frac{4080}{4.186\times27.2}$
$m=\frac{4080}{113.8592}\approx35.8$ g
The mass of the water is approximately 35.8 grams.
# Explanation:
## Step1: Calculate heat lost by granite
$Q_{lost - granite}=m_{granite}c_{granite}\Delta T_{granite}=126.1\times0.795\times40.7$
## Step2: Set up heat - gain equation for water
$Q_{gained - water}=m\times c_{water}\times\Delta T_{water}$
## Step3: Equate heat lost and heat gained
$4080=m\times4.186\times27.2$
## Step4: Solve for mass of water
$m=\frac{4080}{4.186\times27.2}$