Question #02440

Question

Question #02440

1 Answer

Impact of this question

2243 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-27T01:45:57

The water would need to lose $- 1.045 k J$ $-1.045 kJ$ of heat.

Explanation:

In order to determine this value we would use the equation
$Q = m (T_{f} - T_{i}) C_{p})$ $Q = m(T_f-T_i)C_p)$ where

$Q =$ $Q=$ Quantity of heat in Joules
$m =$ $m=$ Mass in grams
$T_{f} =$ $T_f=$ the Final Temp
$T_{i} =$ $T_i=$ the Initial Temp
$C_{p} =$ $C_p=$ the Specific Heat Capacity of the Substance

$Q = ?$ $Q=?$
$m = 225 g$ $m=225 g$
$T_{f} = 10^{o} C$ $T_f=10^oC$
$T_{i} = 25^{o} C$ $T_i=25^oC$
$C_{p} = 4.18 \frac{J}{g^{o} C}$ $C_p=4.18 J/g^oC$ the Specific Heat Capacity of water

$Q = 25 g (10^{o} C - 20^{o} C) 4.18 \frac{J}{g^{o} C}$ $Q = 25g(10^oC-20^oC)4.18 J/g^oC$

$Q = 25 g (- 10^{o} C) 4.18 \frac{J}{g^{o} C}$ $Q = 25g(-10^oC)4.18 J/g^oC$

$Q = - 1045 J$ $Q = -1045 J$

$Q = - 1.045 k J$ $Q = -1.045 kJ$

Science

Math

Humanities

... and beyond

Question #02440

1 Answer

Explanation:

Related questions

Impact of this question