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 try and explain with step by step or show work,

You can use this formula to help:

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:

A. 0.423 cal/g°C

2b2t

explanation:

2b2t

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b