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

`6008` cal

Step-by-step 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