[Heat and calorimeter] I can't get the correct answer

[Waffle24]

Homework Statement

Hey there guys, first of all English is not my main language so it will be a little hard for me to say things in english. I am dutch native speaker so I will translate most of the things in english using google translator. I hope you guys are okay with that.

Task :
One wants the heat capacity of a calorimeter. It does so by one in the empty calorimeter, which has a temperature of 19.3 °C, 50 grams of benzene with a temperature of 13, 5 °C is added. This drops the temperature of the calorimeter to 15, 9 °C.
a)Calculate the value of the heat capacity of this calormeter.
b)Directly after this (the temperature of the calorimeter is still 15.9 °C), a Platinum bullet is added, which has a mass of 15.0 gram, in the calorimeter. This platinum ball has been kept in a glass flame beforehand; the temperature of the calorimeter runs to 23.7 °C . determine the temperature of the gas flame.

Okay, I had no problem answering a.
Now I did the same thing in b, but my answer does not match the one from my book. I tried to look if I missed any $-$ or $+$ sign , but still no luck. I even checked whether I filled the wrong number or not, and to me it seems that everything is correct. The Answer should be 604 °C , but I keep getting below 604 °C. I just want to know if I am doing something wrong or maybe the book answer could be wrong.

Homework Equations

$Q = m.c. \bigtriangleup T$
$Q = C .ΔT$

The Attempt at a Solution

I've uploaded 2 pictures of the calculations I did.
According to my book the answer I got for a is correct.

 

[Chestermiller]

Homework Statement

Hey there guys, first of all English is not my main language so it will be a little hard for me to say things in english. I am dutch native speaker so I will translate most of the things in english using google translator. I hope you guys are okay with
Task :
One wants the heat capacity of a calorimeter. It does so by one in the empty calorimeter, which has a temperature of 19.3 °C, 50 grams of benzene with a temperature of 13, 5 °C is added. This drops the temperature of the calorimeter to 15, 9 °C.
a)Calculate the value of the heat capacity of this calormeter.
b)Directly after this (the temperature of the calorimeter is still 15.9 °C), a Platinum bullet is added, which has a mass of 15.0 gram, in the calorimeter. This platinum ball has been kept in a glass flame beforehand; the temperature of the calorimeter runs to 23.7 °C . determine the temperature of the gas flame.

Okay, I had no problem answering a.
Now I did the same thing in b, but my answer does not match the one from my book. I tried to look if I missed any $-$ or $+$ sign , but still no luck. I even checked whether I filled the wrong number or not, and to me it seems that everything is correct. The Answer should be 604 °C , but I keep getting below 604 °C. I just want to know if I am doing something wrong or maybe the book answer could be wrong.

Homework Equations

$Q = m.c. \bigtriangleup T$
$Q = C .ΔT$

The Attempt at a Solution

I've uploaded 2 pictures of the calculations I did.
According to my book the answer I got for a is correct.

I don't follow your algebra. Where did the 2 come from?

 

[Waffle24]

I don't follow your algebra. Where did the 2 come from?

$15 × 0,13(Tend - Tbegin)$
Tend is given which = 23,7 °C
So it will be $15 × 0,13(23,7 - Tbegin)$
$15 × 0,13 = 1,95$
$1,95(23,7 - Tbegin) = 46,22 - 1,95T$

 

[Chestermiller]

(468+667)/(1.95 )+ 23.7= ??

 

[Waffle24]

(468+667)/(1.95 )+ 23.7= ??

Well the answer = 605,7 , but I'm still confused why you did $+ 23,7$ and not $+46,22$

 

[Chestermiller]

I'm on my iPhone now, so it's hard to do equations. I'll get back with you later when I have my computer. But basically, the heat gained by the benzene and calorimeter is equal to the heat list by the bullet.

 

[Chestermiller]

Heat removed from bullet = $1.95(T-23.7)$. So, $$468+667=1.95(T-23.7)$$

 

[Waffle24]

Ah that explains it, thank you sir for your help.