Question

What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°c to 70.0°c when 2,520 j of heat is applied? use q equals m c subscript p delta t.. 0.00420 j/(gi°c) 0.00661 j/(gi°c) 238 j/(gi°c) 252 j/(gi°c)

Answer 1

Answer:

4.2 j/kg-C      =  .0042 j / (g-C)

Explanation:

Specific heat units are    j / kg-C

   2520 J  / [ (10 kg)(70-10 C)]   = 4.2   j / kg-C

Answer 2

Answer:

0.00420

Explanation:

The equation is Q=mc(T(final)-T(initial), where c is the specific heat, Q is heat supplied, m is mass, T(final) is final temperature and T(initial) is initial temperature (you'll see this written as delta T, which means change in temperature).

2520 = Q

m = 10.0kg; answer choices are in g, not kg, so multiply by 1000 to get m in g; m = 10000 g

Plug in the values you have and solve for c.

2520=(10000)(c)(70-10)

2520=600000c

c=0.0042 j/(gc)