Warm bath from hot and cold water: the mixing calculation

SNC2D Grade 10 Science · Climate Change

A science technician is preparing a 38.0 °C water bath for a class yeast-respiration experiment. The urn holds 2.40 kg of water at 70.0 °C; the cold tap runs at 15.0 °C; the insulated bath tub itself is pre-warmed, so its own uptake can be ignored. The technician wants to know how much tap water to add so the mix lands on 38.0 °C exactly — and, out of curiosity, how long the department's 300 W aquarium heater would have needed to do the same warming job on that tap water alone.

Given
  • m_h = 2.4 kgHot water from the urn
  • T_h = 70 °CUrn temperature
  • T_c = 15 °CCold tap temperature
  • T_mix = 38 °CTarget bath temperature
  • P = 300 WAquarium heater, for the comparison
Determine
  1. (a)the heat the urn water gives up cooling to the 38.0 °C target
  2. (b)the mass of 15.0 °C tap water that heat can raise to the same target
  3. (c)the time the 300 W heater would need for the same warming job
Step 1 of 3(a) · solve for Heat energy

The hot side of the ledger: the urn water's slide is 70.0 − 38.0 = 32.0 K. Keeping each stream's ΔT pinned to its own starting temperature is the whole bookkeeping of mixing — the hot water falls 32, the cold climbs 23, and swapping those two is the error that wrecks this calculation most often.

Rearranged for Q
Q=mcΔTQ = m c \Delta T
Your values, in your units
Q=(2.4 kg)(4,186 J/(kgK))(32 C)Q = \left( 2.4\ \text{kg} \right) \, \left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 32\ \text{C}^{\circ} \right)
Answer
Q=321.48 kJQ = 321.48\ \text{kJ}

Carried onward at full precision, not this rounded figure.

Open the Sensible Heat (Q = mcΔT) solver →

Step 2 of 3(b) · solve for Mass

In an insulated tub, heat lost equals heat gained, so the same Q now funds the cold water's 23.0 K climb: m = Q/(cΔT) ≈ 3.34 kg. Water's c cancels top and bottom, which is why the answer is really just 2.40 × 32/23 — the masses balance in inverse proportion to their temperature swings, a seesaw with the pivot at 38.0 °C.

Rearranged for m
m=QcΔTm = \frac{Q}{c \, \Delta T}
321.48 kJcarried from step 1
Your values, in your units
m=(321,485 J)(4,186 J/(kgK))(23 C)m = \frac{\left( 321{,}485\ \text{J} \right)}{\left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 23\ \text{C}^{\circ} \right)}
Answer
m=3.3391 kgm = 3.3391\ \text{kg}

Carried onward at full precision, not this rounded figure.

Open the Sensible Heat (Q = mcΔT) solver →

Step 3 of 3(c) · solve for Time

The cold water's warming bill IS the hot water's 321.5 kJ — that equality was step 2's whole premise — so the heater comparison is t = E/P: nearly 18 minutes of a 300 W element to do what the pour does in seconds. Mixing moves no heat faster; it just moves heat that was already paid for at the urn.

Rearranged for t
t=EPt = \tfrac{E}{P}
321.48 kJcarried from step 1
Your values, in your units
t=(321,485 J)(300 W)t = \tfrac{\left( 321{,}485\ \text{J} \right)}{\left( 300\ \text{W} \right)}
Answer
t=17.86 mint = 17.86\ \text{min}

Carried onward at full precision, not this rounded figure.

Open the Electrical Energy (E = Pt) solver →

Therefore the urn water surrenders 321.5 kJ on its way down to 38.0 °C, which lifts 3.34 kg of tap water up to the same mark — a 5.74 kg bath — and the 300 W heater would have ground away for 17.9 min to match what one pour accomplished.

Why this order

Mixing problems are double-entry bookkeeping, and the chain keeps the two entries on separate lines on purpose: part (a) prices what the hot stream gives up, part (b) spends exactly that on the cold stream, and the join between them — heat lost equals heat gained — is stated once, where it can be examined. It is an assumption about the TUB (insulated, pre-warmed), not a law of algebra, and labs that skip the pre-warming discover their first batch lands a degree low because the tub itself drank part of the ledger. The recurring student error is crossing the ΔTs: the hot water's 32 belongs to the hot mass, the cold water's 23 to the cold, and because water's specific heat cancels entirely, the surviving structure is a lever balance — m_hot × 32 = m_cold × 23, masses inversely proportional to their swings, with the target temperature as the pivot.

That lever is worth internalizing because it is how every mixing valve in a building thinks: a shower mixer, a boiler's tempering valve, a dairy's pasteurizer-cooler all solve part (b) continuously, in brass. Part (c) reframes the same joules as time and makes the energy tangible: 321.5 kJ is eighteen minutes of a 300 W element, which is why hot-water storage exists at all — a tank is a battery of already-paid-for joules that a pour can withdraw in seconds, at a rate no reasonable element could match live. Change the target to 40 °C and rerun the worksheet: the cold share drops to 2.88 kg, and the seesaw arithmetic shows why the last few degrees of a hotter bath cost disproportionately much cold-side capacity.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.