# 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

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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unett
Principle of calirometry or heat balance states that
Heat lossed=Heat gained
${m}_{1}{c}_{1}\left({T}_{1}-T\right)={m}_{2}{c}_{2}\left(T-{T}_{2}\right)$
${m}_{1}=300grams,{c}_{1}=0.030ca\frac{l}{g}.C,{T}_{1}=100C$
${m}_{2}=100grams,{c}_{2}=1,{T}_{2}=70C$
plug the values you will get the answer 86.81C ˜ the answer is (b)
###### Not exactly what you’re looking for?
Jeffrey Jordon

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

$0.09\left(100-T\right)=T-70$

$9-0.09T=T-70$

$⇒1.09T=79$

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

E)