Question

when 45 g of an alloy, at 25°C, are dropped into 100.0g of water, the alloy absorbs 956J of heat. If the temperature of the alloy is 37°C, what is its specific heat?

A. 0.423 cal/g°C
B. 1.77 cal/g°C
C. 9.88 cal/g°C
D. 48.8 cal/g°C

Please try and explain with step by step or show work, thank you!!

Answer 1

You can use this formula to help:

$c = \frac{q}{m \: \times \: change \: in \: t}$

Where:

C = specific heat

q = heat

m = mass

t = temperature

What we know:

C = unknown

q = 956 J

m = 45 g

change in t = 12°C because 37°C - 25°C = 12°C

Plug known values into the formula:

C = 956 J / (45 g) (12°C) and we are left with a specific heat of 1.77J/g°C

Now, convert Joules to calories and then you get:

Answer: A. 0.423 cal/g°C

Answer 2

Answer:

A. 0.423 cal/g°C

Explanation: