Respuesta :
Answer:
[tex] 6008[/tex] cal
Explanation:
[tex]m_{i}[/tex] = mass of ice = 100 g = 0.1 kg
[tex]c_{i}[/tex] = specific heat of ice = 0.5 cal/(kg°C)
[tex]c_{w}[/tex] = specific heat of water = 1 cal/(kg°C)
[tex]L[/tex] = Latent heat of fusion of ice = 80 J/g
[tex]T_{i}[/tex] = initial temperature of ice = - 20 °C
[tex]T_{f}[/tex] = final temperature of ice = 50 °C
Q = Heat gained
Heat gained is given as
[tex]Q = m_{i} c_{i}(0 - (T_{i}))+ m_{i}L + m_{i}c_{w}(T_{i} - 0)[/tex]
[tex]Q = (100) (0.5) (0 - (- 20))+ (0.1)(80) + (100) (1)(50 - 0)[/tex]
[tex]Q = 6008[/tex] cal
