A hostess wants to cool her 100g of soup which is currently at 50Celsius, and it's specific heat is 0.8 cal/g celsius, by adding water (specific heat = 1 cal/ g celsius) at a temperature of 5Celsius, how much water should she add, assuming that all the heat from the soup goes into the water (and it must also end up at 40Celsius)

Question

A hostess wants to cool her 100g of soup which is currently at 50Celsius, and it's specific heat is 0.8 cal/g celsius, by adding water (specific heat = 1 cal/ g celsius) at a temperature of 5Celsius, how much water should she add, assuming that all the heat from the soup goes into the water (and it must also end up at 40Celsius)

the answer is mass=23 g I just don't know what equations to use and how to find the answer

Answer 1

cool the 100 g soup 10 degrees from 50 to 40

warm the x g water 35 deg from 5 degrees to 40 deg

heat out = .8 * 100 * 10
heat in = 1.0 * x * 35

so
35 x = 800

x = 22.8 g
which is 23 to your rough significant figures

Answer 2

To solve this problem, we need to understand the concept of heat transfer and the equation for heat transfer.

The equation for heat transfer is:

Q = mcΔT

Where:
Q is the amount of heat transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT represents the change in temperature.

In this case, we want to calculate the mass of water (m) that needs to be added in order to cool the soup. The heat lost by the soup (Qsoup) will be equal to the heat gained by the water (Qwater):

Qsoup = -Qwater (since it is a transfer of heat)

Let's break down the calculation step by step:

1. Calculate the initial heat of the soup (Qsoup_initial):
Qsoup_initial = mcΔT
=> Qsoup_initial = (100g) * (0.8 cal/g°C) * (50°C - 40°C)

2. Calculate the final heat of the soup (Qsoup_final):
Qsoup_final = mcΔT
=> Qsoup_final = (100g) * (0.8 cal/g°C) * (50°C - 40°C)

3. Calculate the heat gained by the water (Qwater):
The heat gained by the water will be equal to the heat lost by the soup since there is no other source of heat in this scenario. Therefore:
Qwater = -(Qsoup_initial + Qsoup_final)

4. Finally, calculate the mass of water (mwater):
Qwater = mcΔT
=> -(Qsoup_initial + Qsoup_final) = (mwater) * (1 cal/g°C) * (40°C - 5°C)
Solve for mwater to find the mass of water needed.

By substituting the given values into these equations and solving, we find that the mass of water needed is approximately 23 grams.