How much energy will be required to heat a #"1.0 kg"# mass of water from #25^@"C"# to #99^@"C"#?

Question

Wyatt Adkins · Accepted Answer

You'd need #"310 kJ"# to raise the temperature of 1 kg of water from #25# to #99^@"C"#. That means you have all the necessary data to apply the equation. #q = m * c * DeltaT#, where #q# - the amount of heat needed;
#m# - the mass of water - in your case 1.0 kg:
#c# - the specific heat of water;
#DeltaT# - the difference between the final temperature, #99^@"C"#, and the initial temperature, #25^@"C"#, of the water.  Enter your data into the formula to obtain #q = "1,000 g" * 4.18"J"/("g" * ^@"C") * (99-25)^@"C"# #q = "309,320 J" = "+309.3 kJ"# The response, rounded to two sig figs, is #q = "+310 kJ"#

Noah Aldrich · Answer

It will require #"310,000 J"#.

You must apply the following equation to get the answer to this question: #q = cmDeltaT#, where #q# is the quantity of heat gained or lost, #c# is the specific heat capacity (of water in this case), #m# is mass in grams, and #DeltaT# is the difference in temperature, #DeltaT=T_"final"-T_"initial"# Known/Given: #c_"water"= 4.184 "J"/("g"*""^("o")"C"# #m=1.0color(red)cancel(color(black)("kg"))xx"1000 g"/(1color(red)cancel(color(black)("kg")))=1.0xx10^3# #"g"# #T_i="25"^@"C"# #T_f="99"^@"C"# #DeltaT="99"^@"C" - "25"^@"C"="74"^@"C"# Unknown: #q# Resolution: Enter the known values into the formula and find the solution. #q=4.184"J"/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C")))xx1.0xx10^3color(red)cancel(color(black)("g"))xx74^@color(red)cancel(color(black)("C")) = "310,000 J"# (rounded to two significant figures)

Cooper Anderson · Answer

undefined The specific heat capacity of water is 4.186 J/g°C. The formula to calculate heat energy (Q) is Q = mcΔT, where m is mass, c is specific heat, and ΔT is the temperature change.

Q = (1.0 kg) * (4.186 J/g°C) * (99°C - 25°C). Calculate the result.

Noah Aldrich · Answer

undefined To calculate the energy required to heat a 1.0 kg mass of water from 25°C to 99°C, you can use the formula:

$$ Q = mcΔT $$

Where:
- $ Q $ is the heat energy (in joules),
- $ m $ is the mass of the substance (in kilograms),
- $ c $ is the specific heat capacity of the substance (in joules per kilogram per degree Celsius), and
- $ ΔT $ is the change in temperature (in degrees Celsius).

For water, the specific heat capacity ($ c $) is approximately $ 4186 \, \text{J/kg}^\circ\text{C} $.

Substituting the given values into the formula:

$$ Q = (1.0 \, \text{kg}) \times (4186 \, \text{J/kg}^\circ\text{C}) \times (99^\circ\text{C} - 25^\circ\text{C}) $$

$$ Q = (1.0 \, \text{kg}) \times (4186 \, \text{J/kg}^\circ\text{C}) \times (74^\circ\text{C}) $$

$$ Q = 308,564 \, \text{J} $$

So, the energy required to heat a 1.0 kg mass of water from 25°C to 99°C is $ 308,564 \, \text{J} $.