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

1. 2.510kJ

2. Q = 1.5 kJ

Step-by-step 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*(4.184J)/(1cal)*(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(J)/(g\°C) (34.9\°C-8.5\°C)\\\\Q=1495.3J*(1kJ)/(1000J) \\\\Q=1.5kJ`

Regards!