# Calculate the mass of ice that would melt when 2kg of copper is…

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**Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss.**

**(specific heat capacity of copper = 400Jkg⁻¹k⁻¹, latent heat of fusion of ice = 3.3 x 10⁵Jkg⁻¹)**

**A.**** 8****/****33****kg**

**B.**** 3.3****/****80****kg**

**C.**** 80****/****33****kg**

**D.**** 3.3****/****8****kg**

**QUICK ANSWER…**

**A**

**DETAILS…**

**the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss**** is 8/33 kg. let me explain…**

**heat lost by copper = heat gained by ice**

**mcΔT = ML**

**where:**

**m = mass of copper = 2kg**

**c = ****specific heat capacity of copper = 400Jkg⁻¹k⁻¹**

**ΔT = change in temperature = temperature of boiling water – temperature of ice = 100⁰C – 0⁰C = 373k – 273k = 100k**

**M = mass of ice =?**

**L = ****latent heat of fusion of ice = 3.3 x 10⁵Jkg⁻¹**

**mcΔT = ML**

**therefore,**

**M = mcΔT/L**

**M = (2 × 400 × 100)/(3.3 × 10⁵)**

**M = 80000/(3.3 × 10⁵)**

**M = (8 × 10⁴)/(3.3 × 10⁵), or**

**M = (8 × 10⁴)/(33 × 10⁴)**

**M = (8 × 10⁴)/(33 × 10⁴)**

**M = 8/33kg**

**Now for the right answer to the above question:**

**Option A is correct.****Option B is incorrect.****C is incorrect.****D is not the correct answer.**

**KEY-POINTS…**

**You may please note these/this:**

**Heat lost by hot copper = heat gained by ice.****Or, mcΔT = ML.****In reality, this formula only assumes that all the heat is used to melt the ice only, no heat will be lost to some already melted water in the process or the container.**

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**/ culled from 20****21**** JAMB-UTME ****PHYSICS**** question ****32 ****/**