Question

Calculate the heat gained by 100 grams of ice at -20°C in order to become water at 50°C. ( C = .5 for ice and C = 1 for water, Q = M.С.ΔT)

Answer 1

Answer:

$6008$ cal

Explanation:

$m_{i}$ = mass of ice = 100 g = 0.1 kg

$c_{i}$  = specific heat of ice = 0.5 cal/(kg°C)

$c_{w}$  = specific heat of water = 1 cal/(kg°C)

$L$  = Latent heat of fusion of ice = 80 J/g

$T_{i}$  = initial temperature of ice = - 20 °C

$T_{f}$  = final temperature of ice = 50 °C

Q = Heat gained

Heat gained is given as

$Q = m_{i} c_{i}(0 - (T_{i}))+ m_{i}L + m_{i}c_{w}(T_{i} - 0)$

$Q = (100) (0.5) (0 - (- 20))+ (0.1)(80) + (100) (1)(50 - 0)$

$Q = 6008$ cal