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19 g of water at 30^@C and 5 g of ice at...

19 g of water at `30^@C` and 5 g of ice at `-20^@C` are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice `=0.5calg^(-1) (.^(@)C)^(-1)` and latent heat of fusion of ice `=80calg^(-1)`

A

`0^@C`

B

`-5^@C`

C

`5^@C`

D

`10^@C`

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The correct Answer is:
To solve the problem of finding the final temperature when 19 g of water at 30°C is mixed with 5 g of ice at -20°C, we will follow these steps: ### Step 1: Calculate the heat required to melt the ice The heat required to raise the temperature of the ice from -20°C to 0°C and then to melt it can be calculated using the formula: \[ Q_1 = m \cdot c \cdot \Delta T + m \cdot L_f \] Where: - \( m \) = mass of ice = 5 g - \( c \) = specific heat of ice = 0.5 cal/g°C - \( \Delta T \) = change in temperature = 0°C - (-20°C) = 20°C - \( L_f \) = latent heat of fusion of ice = 80 cal/g Calculating \( Q_1 \): \[ Q_1 = 5 \cdot 0.5 \cdot 20 + 5 \cdot 80 \] \[ Q_1 = 5 \cdot 10 + 400 = 50 + 400 = 450 \text{ cal} \] ### Step 2: Calculate the heat given by the water The heat lost by the water as it cools from 30°C to the final temperature \( T \) can be calculated as: \[ Q_2 = m \cdot c \cdot (T_{initial} - T_{final}) \] Where: - \( m \) = mass of water = 19 g - \( c \) = specific heat of water = 1 cal/g°C - \( T_{initial} \) = 30°C - \( T_{final} \) = \( T \) Calculating \( Q_2 \): \[ Q_2 = 19 \cdot 1 \cdot (30 - T) = 570 - 19T \text{ cal} \] ### Step 3: Set up the heat balance equation Since no heat is lost to the surroundings, the heat gained by the ice must equal the heat lost by the water: \[ Q_2 = Q_1 \] Substituting the values we found: \[ 570 - 19T = 450 \] ### Step 4: Solve for \( T \) Rearranging the equation: \[ 570 - 450 = 19T \] \[ 120 = 19T \] \[ T = \frac{120}{19} \approx 6.32°C \] ### Step 5: Check if all ice melts We need to check if all the ice melts at this temperature. The heat required to melt the ice (450 cal) is less than the heat given by the water (570 cal), confirming that all the ice will melt. ### Final Answer The final temperature of the mixture is approximately **6.32°C**.

To solve the problem of finding the final temperature when 19 g of water at 30°C is mixed with 5 g of ice at -20°C, we will follow these steps: ### Step 1: Calculate the heat required to melt the ice The heat required to raise the temperature of the ice from -20°C to 0°C and then to melt it can be calculated using the formula: \[ Q_1 = m \cdot c \cdot \Delta T + m \cdot L_f \] ...
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CP SINGH-HEAT AND CALORIMETRY-Exercise
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  6. A student takes 50 g wax (specific heat =0.6 kcal//kg^(@)C) and heats ...

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  7. A block of ice at -10^@C is slowly heated and converted to steam at 10...

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  8. A substance of mass M kg requires a power input of P wants to remain i...

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  9. 80 g of water at 30^@C are poured on a large block of ice at 0^@C. The...

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  10. 10 g of ice at 0^@C is mixed with 100 g of water at 50^@C. What is the...

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  11. A piece of ice of mass of 100g and at temperature 0^(@)C is put in 200...

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  12. A 5 g piece of ice at -20^@C is put into 10 g of water at 30^@C. Assum...

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  13. 19 g of water at 30^@C and 5 g of ice at -20^@C are mixed together in ...

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  14. 2kg of ice at 20^@C is mixed with 5kg of water at 20^@C in an insulati...

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  15. 1 kg of ice at 0^@C is mixed with 1 kg of steam at 100^@C (i) The e...

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  16. How much steam at 100^@C will just melt 64 gm of ice at -10^@C?

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  17. Steam at 100^@C is allowed to pass into a vessel containing 10 g of ic...

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  18. Steam is passes into 22 g of water at 20^@C. The mass of water that wi...

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