Question

A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg°C. It contains 250 gm of liquid at 30°C having a specific heat of 0.4 kcal/kg°C. If we drop a piece of ice of mass 10g at 0°C, What will be the temperature of the mixture?

Answer 1

Convert the the grams into calories

1g = 7.716 Cal

100 g = 771.6 cal = 0.7716 kCal

250 g = 1929 cal = 1.929 kCal

10 = 77.16 cal = 0.07716 kCal

Heat capacity of water = 1 kCal / Kg

Latent heat of fusion of ice = 80 kCal / Kg

Heat gained or lost = mass × heat capacity × change in temperature

Since ice is put in the liquid, there is heat loss :

Change in temperature = (30 - T)

Heat lost by calorimeter and liquid = heat gained by ice

Substituting :

(30 - T) 0.7716 × 0.1 + (30 - T) 1.926 × 0.4 = 0.07716 × 80 + T × 0.07716 × 1

2.3148 — 0.07716 T + 23.112 —0.7704T = 6.1728 + 0.07716 T

Collecting like terms together :

0.92472 T = 19.254

T = 20.82° C

Answer 2

Answer:

Explanation:

Convert the the grams into calories

1g = 7.716 Cal

100 g = 771.6 cal = 0.7716 kCal

250 g = 1929 cal = 1.929 kCal

10 = 77.16 cal = 0.07716 kCal

Heat capacity of water = 1 kCal / Kg

Latent heat of fusion of ice = 80 kCal / Kg

Heat gained or lost = mass × heat capacity × change in temperature

Since ice is put in the liquid, there is heat loss :

Change in temperature = (30 - T)

Heat lost by calorimeter and liquid = heat gained by ice

Substituting :

(30 - T) 0.7716 × 0.1 + (30 - T) 1.926 × 0.4 = 0.07716 × 80 + T × 0.07716 × 1

2.3148 — 0.07716 T + 23.112 —0.7704T = 6.1728 + 0.07716 T

Collecting like terms together :

0.92472 T = 19.254

T = 20.82° C