Question

A 50 gram piece of copper at 200°C is placed in 100 grams of water at 25°C. Assuming no loss of heat to the surroundings, determine the final temperature of the water and copper.

Answer #1

**specific heat capacity of copper = 0.385
J/g.oC**

**specific heat capacity of water = 4.184
J/g.oC**

**Let us denote Cu by symbol 1 and water by symbol
2**

**m1 = 50.0 g**

**T1 = 200.0 oC**

**C1 = 0.385 J/goC**

**m2 = 100.0 g**

**T2 = 25.0 oC**

**C2 = 4.184 J/goC**

**T = to be calculated**

**Let the final temperature be T oC**

**use:**

**heat gained by 2 = heat lost by 1**

**m2*C2*(T-T2) = m1*C1*(T1-T)**

**100.0*4.184*(T-25.0) = 50.0*0.385*(200.0-T)**

**418.4*(T-25.0) = 19.25*(200.0-T)**

**418.4*T -10460 = 3850 - 19.25*T)**

**T= 32.7 oC**

**Answer: 32.7 oC**

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