Question

1 kg ice at -10 degree celsius is mixed with 4.4 kg of water at 30 degree Celsius the final temperature of mixture specific heat of ice is 2100 joule per kg

Answer 1

Answer: $T_{f} = 301 K$

Explanation:

Given that,

Mass of ice = 1 kg

Temperature of ice = 283 K

Mass of water = 4.4 kg

Temperature of water = 303 K

Specific heat of ice = 2100 J/kg-K

We know that,

Specific heat of water = 4185.5 J/kg-K

Now, the final temperature of mixture is

$T_{f} = \dfrac{m_{i}T_{i}C_{p}_{i}+m_{w}T_{w}C_{p}_{w}}{m_{i}C_{p}_{i}+m_{w}C_{p}_{w}}$

$T_{f} = \dfrac{1\ kg\times283\ K\times2100\ J/kg-K+4.4\ kg\times303\ K\times4185.5\ J/kg-K}{1\ kg\times2100\ J/kg-K+4.4\ kg\times4185.5\ J/kg-K}$

$T_{f} = 301 K$

Hence, this is the required solution.