The specific heat of water = 4200 J kg⁻¹ K⁻¹ and the latent heat of ice = 3.4 * 10⁵ J kg⁻¹. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams):
The specific heat of water = 4200 J kg⁻¹ K⁻¹ and the latent heat of ice = 3.4 * 10⁵ J kg⁻¹. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams):
Option 1 - <p>64.6</p>
Option 2 - <p>61.7<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>69.3<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>63.8<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
3 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
Answered by
6 months ago
Correct Option - 2
Detailed Solution:
m(L) = m?S?(ΔT)
⇒ m(3.4 * 10?) = (200)(4200)(25)
⇒ m = 61.7
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