Water of mass m2 = 1 kg is contained in a copper calorimeter of mass m1 = 1 kg. Their common temperature t = 100C. Now a piece of ice of mass m3 = 2 kg and temperature is −110C dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper = 0.1 Kcal/ kg0C, specific heat of water = 1 Kcal/kg0C, specific heat of ice = 0.5 Kcal/kg0C, latent heat of fusion of ice = 78.7 Kcal/kg]

Question

Water of mass m2​ = 1 kg is contained in a copper calorimeter of mass m1​  = 1 kg. Their common temperature t = 100C. Now a piece of ice of mass m3​  = 2 kg and temperature is −110C dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper = 0.1 Kcal/ kg0C, specific heat of water = 1 Kcal/kg0C, specific heat of ice = 0.5 Kcal/kg0C, latent heat of fusion of ice = 78.7 Kcal/kg]

Filo · Accepted Answer

Loss in heat from calorimeter + water as temperture changes from

1 0^{0} C

to 0

^{0} C

= m_{1} C_{1} 10 + m_{2} C_{2} 10 =

1110 + 10.110 = 11 k c a

l
Gain in heat of ice as its temperature changes from

- 1 1^{0} C

to 0

^{0} C

$ kcal$
Hence ice and water will coexist at 0

^{0} C

without any phase change

Question Text	Water of mass $m_{2}$ = 1 kg is contained in a copper calorimeter of mass $m_{1}$ = 1 kg. Their common temperature t = $1 0^{0} C$ . Now a piece of ice of mass $m_{3}$ = 2 kg and temperature is $- 1 1^{0} C$ dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper = 0.1 Kcal/ kg $^{0} C$ , specific heat of water = 1 Kcal/kg $^{0} C$ , specific heat of ice = 0.5 Kcal/kg $^{0} C$ , latent heat of fusion of ice = 78.7 Kcal/kg]
Answer Type	Text solution:1
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