Question

1. (8pt) Using dimensional analysis convert 600.0 calories into kilojoules
2. (8pts) how much heat and kilojoules is needed to raise the temperature of 236 g of silver from 8.5°C to 34.9°C? (Cag = 0.24 J/g°C)

Answer 1

Answer:

1. 2.510kJ  

2. Q = 1.5 kJ

Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, we can proceed as follows:

1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

$1cal=4.184J\\\\1kJ=1000J$

Then, we perform the conversion as follows:

$600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ$

2. Here, we use the general heat equation:

$Q=mC(T_2-T_1)$

And we plug in the given mass, specific heat and initial and final temperature to obtain:

$Q=236g*0.24\frac{J}{g\°C} (34.9\°C-8.5\°C)\\\\Q=1495.3J*\frac{1kJ}{1000J} \\\\Q=1.5kJ$

Regards!