# The temperature of 50.0 g of water was raised to 50.0°C by the addition of 1.0 kJ of heat energy. What was the initial temperature of the water?

##### 1 Answer

#### Explanation:

The idea here is that you need to use the mass of water, the heat added to the sample, and the specific heat of water to find the resulting **change in temperature**.

The equation that establishes a relationship between heat absorbed and change in temperature looks like this

#color(blue)(q = m * c * DeltaT)" "# , where

*final temperature* minus *initial temperature*

Plug in your values and solve for **do not** forget to convert the added heat from *kilojoules* to *Joules*

#q = m * c * DeltaT implies DeltaT = q/(m * c)#

#DeltaT = (1.0 * 10^(3)color(red)(cancel(color(black)("J"))))/(50.0color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 4.785^@"C"#

So, if the temperature of the water *changed* by *final temperature* is now

#DeltaT = T_"f" - T_"i" implies T_"i" = T_"f" - DeltaT#

#T_"i" = 50.0^@"C" - 4.785^@"C" = color(green)(45.2^@"C")#