Question

Solve,55.1 grams of tin at 90.6°C is dropped into a calorimeter with 75.4 grams of water at 21.0°C.The temperature of both the metal and the water reaches 27.2°C.Solve for the specific heat of tin. pleaseee once u solve it please explain how u solved it.

Answer 1

Answer:

$C_{tin}=0.560\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, for this calorimetry problem, we can notice that the heat evolved by the hot tin is gained by the cold water as the calorimeter is perfectly isolated, so we can write:

$-Q_{tin}=Q_{water}$

Thus, by defining the heats in terms of mass, specific heat and temperatures, we get:

$-m_{tin}C_{tin}(T_F-T_{tin})=m_{water}C_{water}(T_F-T_{water})$

Now, since we are asked for the specific heat of tin, we solve for it as shown below:

$C_{tin}=\frac{m_{water}C_{water}(T_F-T_{water})}{-m_{tin}(T_F-T_{tin})}$

Thus, when we plug in, we obtain:

$C_{tin}=\frac{75.4g*4.184\frac{J}{g\°C} (27.2\°C-21\°C)}{-55.1g(27.2\°C-90.6\°C)}\\\\C_{tin}=0.560\frac{J}{g\°C}$

Regards!