Answer:
The total amount of heat is 250.64 kcal.
Explanation:
Given that,
Total amount of heat = 2.000 kg
Temperature of snow = -15.00°C
Temperature of water = 38.00°C
Heat of fusion of water = 1.44 kcal/mol
Specific heat of solid water = 0.488 cal/g°C
Specific heat of liquid = 1.00 cal/g°C
We need to calculate the heat required to bring temperature -15°C to 0°C snow
Using formula of heat
![Q_{1}=mc_{p}\Delta T](https://tex.z-dn.net/?f=Q_%7B1%7D%3Dmc_%7Bp%7D%5CDelta%20T)
Put the value into the formula
![Q_{1}=2000\times0.488\times(0+15)](https://tex.z-dn.net/?f=Q_%7B1%7D%3D2000%5Ctimes0.488%5Ctimes%280%2B15%29)
![Q_{1}=14640\ cal](https://tex.z-dn.net/?f=Q_%7B1%7D%3D14640%5C%20cal)
![Q_{1}=14.640\ kcal](https://tex.z-dn.net/?f=Q_%7B1%7D%3D14.640%5C%20kcal)
We need to calculate the heat required to convert 0°C snow into 0°C water
Using formula of heat
![Q_{2}=ml](https://tex.z-dn.net/?f=Q_%7B2%7D%3Dml)
Put the value into the formula
![Q_{2}=\dfrac{2000}{18}\times1.44](https://tex.z-dn.net/?f=Q_%7B2%7D%3D%5Cdfrac%7B2000%7D%7B18%7D%5Ctimes1.44)
![Q_{2}=160\ kcal](https://tex.z-dn.net/?f=Q_%7B2%7D%3D160%5C%20kcal)
We need to calculate heat required to increases temperature of water from 0°C to 38°C
Using formula of heat
![Q_{3}=mc_{p}\Delta T](https://tex.z-dn.net/?f=Q_%7B3%7D%3Dmc_%7Bp%7D%5CDelta%20T)
Put the value into the formula
![Q_{3}=2000\times1.00\times(38-0)](https://tex.z-dn.net/?f=Q_%7B3%7D%3D2000%5Ctimes1.00%5Ctimes%2838-0%29)
![Q_{3}=76000\ cal](https://tex.z-dn.net/?f=Q_%7B3%7D%3D76000%5C%20cal)
![Q_{3}=76\ kcal](https://tex.z-dn.net/?f=Q_%7B3%7D%3D76%5C%20kcal)
We need to calculate the total heat
Using all heat amount
![Q=Q_{1}+Q_{2}+Q_{3}](https://tex.z-dn.net/?f=Q%3DQ_%7B1%7D%2BQ_%7B2%7D%2BQ_%7B3%7D)
Put the value into the formula
![Q=14.640+160+76](https://tex.z-dn.net/?f=Q%3D14.640%2B160%2B76)
![Q=250.64\ kcal](https://tex.z-dn.net/?f=Q%3D250.64%5C%20kcal)
Hence, The total amount of heat is 250.64 kcal.