# Question 85239

Oct 5, 2017

Pretty Huuugeee!

An edited version was added below which yielded an answer of ${57.143}^{\circ} K$.

#### Explanation:

The question provided a value of specific heat capacity with incorrect units. In the answer immediately below, Gaya3 made an assumption that a value of $C = 3.5 {\text{J""kg"^-1"K}}^{-} 1$ was intended. The answer, using that value for C, was 57143K.

I, Steve J, am making a different interpretation of what was intended for C. I have added Answer 2 , below Gaya3's work (and using that format for the math), a solution using $C = 3.5 {\text{kJ""kg"^-1"K}}^{-} 1$. The difference being that the J is changed to kJ.

q=mC∆T

Here,

$q = 2 M J = 2 \times {10}^{6} J$
$m = 10 k g$
$C = 3.5 {\text{J""kg"^-1"K}}^{-} 1$
∆T=?

So, 2times10^6J=10cancel(kg) times3.5 "J"cancel("kg"^-1)"K"^-1times∆T

=>∆T=(2times10^6cancel(J))/(10 times 3.5 cancel("J")"K"^-1)~~57143K

q=mC∆T

Here,

$q = 2 M J = 2 \times {10}^{6} J$
$m = 10 k g$
$C = 3.5 {\text{ kJ""kg"^-1"K"^-1 = 3.5 times10^3"J""kg"^-1"K}}^{-} 1$
∆T=?

So, 2times10^6J=10cancel(kg) times3.5 times 10^3"J"cancel("kg"^-1)"K"^-1times∆T

=>∆T=(2times10^6cancel(J))/(10 times 3.5 times 10^3cancel("J")"K"^-1)~~57.143^@K#