The specific heat of lead is 0.030 cal/g.C. 300 grams of leadshot at 100C is mixed with 100 grams of water at 70C in an insulated container. The final temperature of the mixture is: A) 100C b) 85.5C c)79.5C d) 74.5C e)72.5C

Question

The specific heat of lead is 0.030 cal/g.C. 300 grams of leadshot at 100C is mixed with 100 grams of water at 70C in an insulated container. The final temperature of the mixture is:
A) 100C
b) 85.5C
c)79.5C
d) 74.5C
e)72.5C

Answer 1

Principle of calirometry or heat balance states that
Heat lossed=Heat gained

m_{1} c_{1} (T_{1} - T) = m_{2} c_{2} (T - T_{2})

m_{1} = 300 g r a m s, c_{1} = 0.030 c a \frac{l}{g} . C, T_{1} = 100 C

m_{2} = 100 g r a m s, c_{2} = 1, T_{2} = 70 C

plug the values you will get the answer 86.81C ˜ the answer is (b)

Answer 2

$300 \cdot 0.03 (100 - T) = 100 \cdot 1 \cdot (T - 70)$

$0.09 (100 - T) = T - 70$

$9 - 0.09 T = T - 70$

$\Rightarrow 1.09 T = 79$

$T = \frac{79}{1.09} \approx {72.47}^{\circ} C$

E)

The specific heat of lead is 0.030 cal/g.C. 300 grams of leadshot at 100C is mixed with 100 grams of water at 70C in an insulated container. The final temperature of the mixture is: A) 100C b) 85.5C c)79.5C d) 74.5C e)72.5C

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